mole concept

PHYSICAL CHEMISTRY By-- P.K.SINHA

Contact at [email protected]

Moles Concept, What is a Mole?

Since atoms and molecules are so tiny, it is convenient to talk about a large number of them at a time. The chemical counting unit is known as the mole. A mole is defined as the amount of substance that contains as many elementary entities (atoms, molecules, or other particles) as there are atoms in exactly 12 g of the 6

12C isotope. It has been found experimentally that

1 mole of particles = 6.022 x 1023 particlesThis value is known as Avogadro�s number. Just like 1 dozen of anything always contains 12 items, 1 mole of anything always contains 6.022 x 1023 items.

Atomic Mass and Molar Mass Isotopic masses cannot be obtained by summing the masses of the elementary particles (neutrons, protons, and electrons) from which the isotope is formed. This process would give masses slightly too large, since mass is lost when the neutrons and protons come together to form the nucleus. Atomic masses (also called atomic weights) are thus assigned relative to the mass of a particular carbon isotope, 6

12C , which is assigned the mass of 12 amu exactly. Likewise 1 mole of 612C has a mass of

exactly 12 g. Atomic masses and molar masses of other isotopes are calculated based on their mass relative to that of Carbon-12. Masses of �average� atoms are found by summing isotopic masses, weighting each isotopic mass by its abundance . Thus one �average� C atom has a mass of 12.01 amu, and the mass of 1 mole of �average� carbon atoms has a mass of 12.01 g. These average masses are what are given on the periodic chart.

Molecular Masses and Compound Masses Molecular masses are found by summing atomic masses . They are often called molecular weights. Thus the mass of 1 mole of water, H2O, would be 2 x (molar mass of H) plus 1x (molar mass of O) or [(2 x 1.008 g) + (1 x 16.00 g)] = 18.02 g. Ionic compounds such as NaCl do not contain molecules. Their formulas give the relative numbers of each kind of atom in the sample. What we mean by the molar mass (or the molecular weight) of an ionic compound is really the formula weight. The formula weight is the sum of the atomic masses in the formula.

Percent Composition of Compounds The percent composition by mass is the percent by mass of each element in a compound. If there are n moles of an element per mole of compound, the percent by mass of the element is calculated using the equation,

% Composition of Element =n molar mass of element

molar mass of compound

100%

The sum of the % compositions of all elements in a compound is 100%. Exercises

1. The atomic mass scale gives masses in atomic mass units (amu) relative to the mass of carbon-12. (a) What is the mass of one 12C atom in atomic mass units (amu)? (b) What is the mass of an average C atom in atomic mass units (amu)? (c) What is the mass of an average Cl atom in amu? (d) What is the mass of an average Br atom in amu? 2. The molar mass scale gives masses in grams (g) relative to the mass of 12C. (a) What is the mass in grams of 1 mole (mol) of 12C? (b) What is the mass in grams of 1 mole (mol) of carbon? (c) What is the mass in grams of 1 mole (mol) of Cl? (d) What is the mass in grams of 1 mole (mol) of Na? 3. How many 12C atoms are present in a mole of 12C ? 4. Cinnamic alcohol is used mainly in perfumery, particularly in soaps and cosmetics. Its molecular formula is C9H10O. (a) Calculate the percent composition by mass of C, H, and O in cinnamic alcohol. (b) How many molecules of cinnamic alcohol are contained in a sample of mass 0.469 g?

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Answers: 1. (a) 12 amu exactly; (b) 12.01 amu; (c) 35.45 amu; (d) 79.90 amu. 2. (a) 12 g exactly; (b) 12.01 g; (c) 35.45 g; (d) 22.99 g. 3. 6.022 x 1023 atoms of 12C. 4. (a) 80.56% C; 7.51% H; 11.93% O; (b) 2.11 x 1021 molecules of C9H10O.

Empirical and Molecular Formulas The empirical formula of a compound gives the simplest whole number ratio of different types of atoms in the compound. All salt formulas are empirical formulas. On the other hand, the molecular formula of a compound may or may not be the same as its empirical formula. For example, the molecular formula of butane is C4H10 while its empirical formula is C2H5. The molecular formula gives the true number of each kind of atom in a molecule.

Empirical formulas may be easily determined from experimental data.

Usually you must first determine how many grams of each type of atom are in the compound. If percent composition data is given, assume that you have 100.0 g of the compound; then the number of grams of each element is equal to the percentage for that element.

The next task is convert the grams of each element to moles of the element. Be sure to keep at least three significant figures in your answers.

The final step is to write the molar amounts of each element as subscripts in the formula. Then divide all molar subscripts by the smallest value in the set. At this point, the subscripts may all be very close to whole numbers; if so, you are finished. If one (or more) of the subscripts is not close to a whole number, multiply all molar subscripts by the simple factor which makes all subscripts whole numbers.

Once the empirical formula is determined, the molecular formula is easily found if the molar mass (molecular weight) of the molecule is also known. You first calculate the molar mass of the empirical formula. Then you divide the molar mass of the molecule by the molar mass of the empirical formula. The division should give a simple whole number. That number is the factor by which all subscripts in the empirical formula must be multiplied to obtain the molecular formula.

Exercises

1. The molecular formula of the antifreeze ethylene glycol is C2H6O2. What is the empirical formula?

2. A well-known reagent in analytical chemistry, dimethylglyoxime, has the empirical formula C2H4NO. If its molar mass is 116.1 g/mol, what is the molecular formula of the compound?

3. Nitrogen and oxygen form an extensive series of oxides with the general formula NxOy. One of them is a blue solid that comes apart, reversibly, in the gas phase. It contains 36.84% N. What is the empirical formula of this oxide?

4. A sample of indium chloride weighing 0.5000 g is found to contain 0.2404 g of chlorine What is the empirical formula of the indium compound?

Answers: 1. CH3O 2. Molar mass of empirical formula is 58.06 g/mol. Thus molecular formula is C4H8N2O2. 3. The ratios are N1.00O1.50 . Since 1.50 is not close to a whole number, we multiply both subscripts by 2. The empirical formula is thus N2O3. (The name is dinitrogen trioxide.) 4. InCl3.

Chemical Stoichiometry Problems Calculating the yield of a chemical reaction is a process at the heart of chemistry. While there are many ways a problem can be phrased, in all cases the stoichiometric coefficients in the balanced reaction are used to determine the mole ratios between reactants and products. Thus the first step is usually calculating the moles of each species available. If an amount is given in grams, the molar mass is used

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as a conversion factor to change grams to moles.

Limiting Reagent Problems In some problems, amounts of more than one species are given. In that case your first task is to determine which species is the limiting reagent. Just as you can make only 1 bicycle from 2 wheels and 4 handlebars (with 3 handlebars left over), and only 2 bicycles from 8 wheels and 2 handlebars (with 4 wheels left over), in chemical reactions some species are limiting while others may be present in excess.

In the case of a bicycle, we need 2 wheels

1 handlebar

. We obtain analogous information about the relative

amounts of species that react from the stoichiometric coefficients in a balanced chemical equation. For example, in Exercise (2) below the equation CO(g) + 2 H2(g) CH3OH (l)

tells us we need 2 mol H2

1 mol CO

. If we have more than 2 moles of H2 for each mole of CO, CO will be the

limiting reagent and the excess H2 will not react. Conversely, if we have more than 1 mole of CO for every 2 moles of H2, H2 will be the limiting reagent and the excess CO(g) will be left over. In each case, the yield of CH3OH is determined by the moles of limiting reagent available.

Calculating the Theoretical Yield The theoretical (maximum possible) yield is based on the amount of limiting reagent available. The yield is calculated in steps:

�Calculate moles of all reactants available. If amounts are given in grams, convert grams to moles using

the molar mass of each reactant as your conversion factor: 1 mole reactant

# g reactant

.

�NOTE: Skip this step if you have already identified the limiting reagent. To determine which reagent is limiting, use the mole ratio obtained from the balanced equation for the reaction to find the moles of reactant B needed to react with the available moles of reactant A. If the moles of B available are less than the moles of B needed, reactant B is the limiting reagent and reactant A is in excess. Conversely, if the moles of B available are more than the moles of B needed, A is the limiting reagent and B is in excess.

�Calculate the moles of product based on the moles of limiting reagent available; use the stoichiometric

ratio of # moles product

# moles limiting reagent

as the conversion factor.

�If you are asked for the yield in grams, convert the yield in moles to a yield in grams using the molar

mass as your conversion factor: # g product

1 mole product

Percent Yield Most reactions do not go to completion, and so the actual yield is less than the percent yield. The percent yield is calculated as

Percent yield =actual yield

theoretical yield

100%

Exercises: 1. Ammonia is produced by the reaction 3 H2(g) + N2(g) 2 NH3(g) (a) If N2(g) is present in excess and 55.6 g of H2(g) reacts, what is the theoretical yield of NH3(g)? (b) What is the percent yield if the actual yield of the reaction is 159 g of NH3(g)? Answers: 1(a) 313 g NH3(g); (b) 50.8% yield. 2. Methyl alcohol (wood alcohol), CH3OH, is produced via the reaction CO(g) + 2 H2(g) CH3OH (l) Answers: 2(a) CO is the limiting reagent; (b) 8.52 g CH3OH; (c) 0.13 g H2; (d) 88.3%

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Limiting Reagent Problems Limiting reagent problems are simple to recognize as the initial amounts of more than one species will be given. For instance, for the reaction:

2H2(g) + O2(g) 2H2O If the amounts of both H2 and O2 are given, then this is a limiting reagent problem; otherwise, this is not a limiting reagent problem. There are 6 steps to a limiting reagent problem: 1. Convert the amounts of all species to moles 2. Divide the number of moles of each reactant species by its stoichiometric coefficient. The smallest number corresponds to the limiting reagent. 3. Determine the moles of all other species that react or form by multiplying the smallest number from step 2 by each species' stoichiometric coefficient. 4. Add/Subtract the moles determined in step 3 to/from the initial moles. 5. Convert moles to amounts 6. Check to make sure that total initial mass = total final mass Problem-1..In the reaction Mg3N2 + 6H2O 3Mg(OH)2 + 2NH3, if the initial amount of Mg3N2 is 58.1g and the initial amount of H2O is 20.4g, what are the final masses of each specie? 1. Convert the amounts of all species to moles MM(Mg3N2) = 100.93 g Mg3N2 / mol Mg3N2 mass(Mg3N2) = 58.1 g Mg3N2 x (1 mol Mg3N2 / 100.93 g Mg3N2) = 0.576 mol Mg3N2 MM(H2O) = 18.02 g H2O / mol H2O mass(H2O) =20.4 g H2O x (1 mol H2O / 18.02 g H2O) = 1.13 mol H2O

Compound Mg3N2 H2O Mg(OH)2 NH3

MM 100.93 18.02 58.32 17.03

Initial Mass 58.1 g 20.4 g 0.0 g 0.0 g

Initial Moles 0.576 mol Mg3N2

1.13 mol H2O 0.0 mol Mg(OH)2

0.0 mol NH3

1. Divide the number of moles of each reactant species by its stoichiometric coefficient. The smallest amount is the limiting reagent.

Step 2 Division 0.576 mol / 1 = 0.576 mol

1.13 mol / 6 = 0.189 mol� SMALLEST NUMBER SO LIMITING REAGENT

1. Determine the amount of all other species that react or form by multiplying the smallest number from step 2 by each species stoichiometric coefficient.

Step 2 Division 0.576 mol / 1 = 0.576 mol


Step 3 Amounts 1 x 0.189 mol = 0.189 mol

Mg3N2

6 x 0.189 mol = 1.13 mol

H2O

3 x 0.189 mol = 0.566 mol

Mg(OH)2

2 x 0.189 mol = 0.377 mol NH3

1. Add/Subtract the amounts determined in step 3 to/from the initial amounts.

Final Moles

0.576 mol - 0.189 mol = 0.387 mol

Mg3N2

1.13 mol - 1.13 mol =

0.0 mol H2O

0.0 mol + 0.566 mol = 0.566 mol Mg(OH)2

0.0 mol + 0.377 mol = 0.377 mol

NH3

1. Convert moles to amounts

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Final Mass

0.387 mol x 100.93 g/mol

= 39.1 g Mg3N2

0.0 g

0.566 mol x 58.32 g/mol =

33.0 g Mg(OH)2

0.377 mol x 17.03 g/mol =

6.42 g NH3

1. Check to make sure that total initial mass = total final mass 2. + 20.4 = 78.5; 39.1 + 33.0 + 6.42 = 78.5�Mass is conserved! Problem 2.-In the reaction Au2S3 + 3H2 3H2S + 2Au, if the initial amount of Au2S3 is 500.20 g and the initial amount of H2 is 5.67g, what are the final masses of each specie? 1. Convert the amounts of all species to moles MM(Au2S3) =490.15 g Au2S3 / mol Au2S3 mass(Au2S3) =500.20 g Au2S3x (1 mol Au2S3/ 490.15 g Au2S3) = 0.1.0204 mol Au2S3 MM(H2) = 2.02 g H2/ mol H2 mass(H2) =5.67 g H2x (1 mol H2/ 2.02 g H2) = 2.8125 mol H2

Compound Au2S3 H2 H2S Au

MM 490.15 2.02 34.09 196.97

Initial Mass 500.20 g 5.67 g 0.0 g 0.0 g

Initial Moles 1.0204 mol Au2S3

2.8125 mol H2 0.0 mol H2S 0.0 mol Au


Step 2 Division 1.0204 mol / 1 = 1.0204

mol



Step 3 Amounts 1 x 0.9375

mol = 0.9375 mol Au2S3

3 x 0.9375 mol = 2.8125

mol H2

3 x 0.9375 mol = 2.8125

mol H2S

2 x 0.9375 mol = 1.875 mol Au


Final Moles

1.0204 mol - 0.9375 mol = 0.08288 mol

Au2S3

2.8125 mol - 2.8125 mol =

0.0 mol H2

0.0 mol + 2.8125 mol = 2.8125 mol

H2S

0.0 mol + 1.875 mol = 1.875 mol

Au


Final Mass

0.08288 mol x 490.15

g/mol = 40.6 g Au2S3

0.0 g 2.8125 mol x 34.09 g/mol =

95.9 g H2S

1.875 mol x 196.97 g/mol =

369.4 g Au

1. Check to make sure that total initial mass = total final mass 500.20 + 5.67 = 505.87; 40.6 + 95.9 + 369.4 = 505.9�Mass is conserved! Problem-3-In the reaction 2C3H6 + 2NH3 + 3O2 2C3H3N + 6H2O, if the initial amounts are C3H6 22.5g, NH3 20.6 g, and O2 18.1 g, what are the final amounts of each specie? 1. Convert the amounts of all species to moles

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Compound C3H6 NH3 O2 C3H3N H2O

MM 42.078 17.034 32.000 53.064 196.97

Initial Mass 22.5 g 20.6 g 18.1 g 0.0 g 0.0 g

Initial Moles 0.5347 mol C3H6

1.2094 mol NH3

0.5656 mol O2

0.0 mol C3H3N

0.0 mol H2O


Step 2 Division

0.5347 mol / 2 =0.2675

mol

1.2094 mol / 2 = 0.6047

mol



Step 3 Amounts

2 x 0.1885 mol =

0.3771 mol C3H6

2 x 0.1885 mol =

0.3771 mol NH3

3 x 0.1885 mol =

0.5656 mol O2

2 x 0.1885 mol =

0.3771 mol C3H3N

6 x 0.1885 mol

=1.1313 mol H2O


Final Moles

0.5347 - 0.3771 =

0.1576 mol C3H6

1.2094 - 0.3771 =

0.8323 mol NH3

0.5656 - 0.5656 = 0.0

mol O2

0.0 + 0.3771 = 0.3771

mol C3H3N

0.0 + 1.1313 =

1.1313 mol H2O

Final Mass


Final Mass

0.1576 mol x (42.078 g

/mol) = 6.6 g C3H6

0.8323 mol x (17.034 g/mol) =

14.2 g NH3

0.0 g O2

0.3771 x (53.064 g/mol) = 20.0 g C3H3N

1.1313 x (18.016 g/mol) =

20.4 g H2O

1. Check to make sure that total initial mass = total final mass 22.5 + 20.6 + 18.1 = 61.2; 6.6 + 14.2 + 20.0 +20.4 = 61.2g�Mass is conserved! Now why do steps 2 and 3 give the correct results? Give a reaction aA +bB cC + dD, we know that a moles of A will react with b moles of B to give c moles of C and d moles of D. This means, for instance, that Y moles of A will require Y mol A x (b mol B / a mol A) = Yb/a mol B, since the stoichiometric coefficients are conversion factors between different species. We define 1 mol of this reaction as the process of mixing a mol of A with b mol of B to form c mol C and d mol D. Thus, if we have Y mol of A, we can convert to mol of reaction by Y x (1 mol reaction / a mol A) = Y/a. Thus, the divisions in step 2 convert each initial amount into units of mol reaction and, hence, calculate the amount of the reaction that each reactant can permit. The smallest of these amounts is then the least amount of reaction that any initial amount of a reactant will allow. Step 3 is then converting from units of mol reaction to mol of a species by multiplying by Z mol reaction x (c mol C / 1 mol reaction), .

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Rules for Working Limiting Reagent Problems Determine the number of moles of each reactant.

Find the number of moles of desired product that can be produced by each reactant.

The limiting reagent is the one that produces the smallest amount of that product.

The amount of product produced by that limiting reagent is the final answer. Convert to grams if necessary.

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Problem Solving In Solutions Concentrations of Solutions There are a number of ways to express the relative amounts of solute and solvent in a solution. This page describes calculations for four different units used to express concentration: Percent Composition (by mass) Molarity Molality Mole Fraction

Percent Composition (by mass) We can consider percent by mass (or weight percent, as it is sometimes called) in two ways: The parts of solute per 100 parts of solution. The fraction of a solute in a solution multiplied by 100. We need two pieces of information to calculate the percent by mass of a solute in a solution: The mass of the solute in the solution. The mass of the solution. Use the following equation to calculate percent by mass:

Molarity

Molarity tells us the number of moles of solute in exactly one liter of a solution. (Note that molarity is spelled with an "r" and is represented by a capital M.) We need two pieces of information to calculate the molarity of a solute in a solution: The moles of solute present in the solution. The volume of solution (in liters) containing the solute. To calculate molarity we use the equation:

Molality

Molality, m, tells us the number of moles of solute dissolved in exactly one kilogram of solvent. (Note that molality is spelled with two "l"'s and represented by a lower case m.) We need two pieces of information to calculate the molality of a solute in a solution: The moles of solute present in the solution. The mass of solvent (in kilograms) in the solution. To calculate molality we use the equation:

Mole Fraction

The mole fraction, X, of a component in a solution is the ratio of the number of moles of that component to the total number of moles of all components in the solution. To calculate mole fraction, we need to know: The number of moles of each component present in the solution. The mole fraction of A, XA, in a solution consisting of A, B, C, ... is calculated using the

equation: To calculate the mole fraction of B, XB, use:

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MOLE AND PERCENT WORKSHEET . 1. C2H2 is the molecular formula for ethylyne ( acetylene). (a) How many atoms are in one molecule? (b) Which atoms make up acetylene? (c) How many moles of atoms are in one molecule of acetylene? (d) How many molecules are in 5.3 moles of acetylene? (e) How many atoms are in a mole of acetylene? 2. Calculate the molar mass of a mole of the following materials: (a) Al (b) Ra (c) Co (d) CO (e) CO2 (f) HCl (g) Na2CO3 (h) Ca(NO3)2 (i) (NH4)3(PO4) (j) H2O (k) Epsom salts - Mg(SO4)·7H2O (m) blue vitriol - Cu(SO4)·5H2O ? 3. Calculate the number of moles in: (a) 2.3 # of carbon (b) 0.014 g of Tin (c) a 5 Oz silver bracelet (d) a pound of table salt (e) a 350 kg cast iron engine block (f) a gal. of water (8.3 #) (g) a ton of sand (SiO2) (h) 6.2 grams of blue vitriol (i) a pound of Epsom salts ? 4. Calculate the number of atoms in: (a) 100 g of Argon (b) 1.21 kg aluminum foil (c) a 28 # lead brick (d) the E7 kg of water in an olympic swimming pool (e) 7 kg of hydrogen gas (f) a tonne of calcium nitrate ? 5. What is the percentage composition of oxygen in each of the following materials: (a) CO (b) CO2 (c) (NO3)

- (d) isopropyl alcohol C3H8O (e) calcium nitrate (f) blue vitriol - Cu(SO4)·5H2O ? 6. What is the percentage composition of phosphate in each of the following materials: (a) phosphoric acid 7. What is the percentage composition of sulfate in each of the following materials: (a) sulfuric acid (b) sodium sulfate (c) Epsom salts ( d) aluminum sulfate ?

ANSWERS 1a. 4 b. C & H c. 6.64 E-24 d. 3.1922 E24 e. 2.4092 E24

2a. 27.0 b. 226.0 c. 58.9 d. 28.0 e. 44.0

f. 36.5 g. 106.0 h. 164.1 i. 149.0 j. 18.0

k. 246.4 m. 249.6 3a. 86.9 b. 1.18 E-4 c. 1.31

d. 7.75 e. 6.27 E3 f. 210 g. 1.51 E4 h. 0.0248

i. 1.84 4a. 1.51 E24 b. 2.69 E25 c. 3.69 E25 d. 1.00 E33 e. 4.22E27 f. 3.30 E28 5a. 57.1% b. 72.7% c. 77.4%

d. 26.7% e. 58.5% f. 57.7% 6a. 96.9% b. 0%

c. 63.8% d. 61.2% 7a. 98.0% b. 67.6% c. 39.0%

d. 84.2%

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Multiple Choice Question Set 1 1. The density of chlorine relative to air is

(a) 2.44 (b) 3 (c) found experimentally (d) 4

2. A gaseous oxide contains 30.4% of nitrogen, one molecule of which contains one nitrogen atom.

The density of the oxide relative to oxygen is

(a) 0.94 (b) 1.44 (c) 1.50 (d) 3.0

3. The mass of an oxygen atom is half that of a sulphur atom. Can we decide on this basis that the

density of sulphur vapour relative to oxygen is 2?

(a) Yes (b) No

4. Density of air is 0.001293 g/cc. Its vapour density is

(a) 0.001293 (b) 1.293 (c) 14.48 (d) cannot be calculated

5. 5.6 litres of oxygen at NTP is equivalent to

(a) 1 mole (b) ½ mole (c) 1/4 mole (d) 1/8 mole

6. 22.4 litres of water vapour at NTP, when condensed to water, occupies an approximate volume of

(a) 18 litres (b) 1 litre (c) 1 mL (d) 18 mL

7. Which of the following has the highest mass ?

(a) 1g-atom of C (b) 1/2 mole of CH4 (c) 10 mL of water (d) 3.011 × 1023 atoms of O

8. 6.022 × 1022 molecules of N2 at NTP will occupy a volume of

(a) 22.4 litres (b) 2.24 litres (c) 6.02 litres (d) 6.02 mL

9. How many grams are contained in 1 gram-atom of Na ?

(a) 13 g (b) 23 g (c) 1g (d) 1/23 g

10. The weight of 350 mL of a diatomic gas at 0°C and 2 atm pressure 1 g. The wt. of one atom is

(a)

19108 6 022

108

. .

16N (b)

32N (c) 16N (d) 32N

11. The number of atoms present in 16 g of oxygen is

(a) 6.02 × 1011.5 (b) 3.01 × 1023 (c) 3.01 × 1011.5 (d) 6.02 × 1023

12. 1 mole of a compound contains 1 mole of C and 2 moles of O. The moleculr wt of the compound is

(a) 3 (b) 12 (c) 32 (d) 44

13. The volume of a gas at 0°C and 700 mm pressure is 760 cc. The number of molecules present in

this volume is

(a) 1.88 × 1022 (b) 6.022 × 1023 (c) 18.8 × 1023 (d) 18.8 × 1022

14. 1 mole of a diatomic element X2 contains 34 and 40 moles of electrons and neutrons respectively.

The isotopic formula of the element is

(a) 3474 X (b) 17

37 X (c) 3440 X (d) 20

40 X

15. 2 moles of H atoms at NTP occupy a volume of

(a) 11.2 litres (b) 44.8 litres (c) 2 litres (d) 22.4 litres

16. No. of electrons in 1.8 mL of H2O(l) is

(a) 6.02 × 1023 (b) 3.011 × 1023 (c) 0.6022 × 1023 (d) 60.22 × 1023

17. Moleculear weight of a gas, 11.2 litres of which at NTP weighs 14 g. is

(a) 14 (b) 28 (c)

14112. (d) 14 × 11.2

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18. The weight of 1 mole of a gas of density 0.1784 g/L at NTP is

(a) 0.1784g (b) 1 g (c) 4 g (d) cannot be calculated

19. Number of HCl molecules preent in 10 mL of 0.1 N HCl solution is

(a) 6.022 × 1023 (b) 6.022 × 1022 (c) 6.022 × 1021 (d) 6.022 × 1020

20. Number of atoms in 12 gm of 612 C is

(a) 6 (b) 12 (c) 6.022 × 1023 (d) 12 × 6.022 × 1023

21. 5 moles of a gas in a closed vessel was heated from 300 K to 600 K. The pressure of the gas

doubled. The numbe of moles of the gas will be

(a) 5 (b) 2.5 (c) 10 (d) 20

22. Which of the following contains the greatest number of oxygen atoms ?

(a) 1g of O (b) 1g of O2 (c) 1g of O3 (d) All have the same no of atoms

23. If the atomic weight of carbon were set at 24 amu, the value of the Avogadro constant would be

(a) 6.022 × 1023 (b) 12.044 × 1023 S (c) 1 × 1023 s (d) none of these

24. 24If 32 g of O2 contains 6.022 × 1023 molecules at NTP then 32 g of S, under the same conditions,

will contain,

(a) 6.022 × 1023 S (b) 3.011 × 1023 S (c) 12.044 × 1023S (d) 1 × 1023 S

25. How many moles of electrons weigh one kilogram ?

(a) 6.022 × 1023 (b)

19108

1031

.

(c) 6 0229108

1054..

(d)

19108 6 022

108

. .

Answer To Multiple Choice Question Set 1

1 2 3 4 5 6 7 8 9 10 a b b C C d a b b a 11 12 13 14 15 16 17 18 19 20 d d a B D a b c d c 21 22 23 24 25 a d b A D

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Multiple Choice Question Set 2 1. If 16 grams of O2 react with excess C2H6, how many grams of CO2 will be formed? The formula mass of O2 = 32 amu and the formula mass of CO2 = 44 amu. The balanced chemical equation is...

(a) 22 grams (b)13 grams (c)9 grams (d)7 grams 2. If 64 grams of O2 react with 45 grams of C2H6, how many grams of CO2 will be formed? The formula mass of O2 = 32 amu, C2H6 = 30 and the formula mass of CO2 = 44 amu. The balanced chemical equation is given in the previous problem. (a) 132 grams (b)66 grams (c)50 grams (d)44 grams 3. How many milliliters of 2 M NaCl solution are required to make 1 liter of 0.4 M NaCl solution? (a) 5,000 mL (b)800 mL (c)200 mL (d)0.2 mL 4. If an air bag in a car needs 44.8 L of gas when filled, how many grams of NaN3 are needed to fill the bag with nitrogen gas at standard temperature and pressure? Remember that one mole of any gas will occupy 22.4 L at STP. The balanced chemical reaction is...

(a) 56 grams (b)87 grams (c)130 grams (d)1.3 grams 5. How many grams of NaOH are needed to make 100 milliliters of a 0.2 molar solution of NaOH? (a) 0.02 grams (b)0.8 grams (c)20 grams (d)800 grams 6. How many grams of NaHCO3 (baking soda) would you need to neutralize 500 mL of battery acid (H2SO4) that has been spilled on your garage floor? Assume that the concentration of the battery acid is 12 molar. The balanced chemical equation is...

(a) 1,008,000 grams (b)1,008 grams (c)504 grams (d)252 grams 7. When 157.0 grams of CaSO4 are dissolved in enough water to yield a volume of 7.25x102 milliliters of solution. The molarity of this solution will be... (a) 0.0016 M (b)0.837 M (c)1.15 M (d)1.59 M 8. How many grams of carbon are needed to completely react with 75.2 g of SiO2? The balanced

chemical equation is... (a) 3.76 g (b)15.1 g (c)36.0 g (d)45.1 g 9. How many milliliters of 4.00 M NaOH must be added to 100.0 mL of 0.2000 M H2SO4 solution to completely neutralize the acid? The balanced chemical equation is...

(a) 0.04 mL (b)5.0 mL (c)10.0 mL (d)500 mL 10. In an acid-base titration, 42.90 mL of 0.825 M H2SO4 was required to neutralize 75.0 mL of NaOH solution. What is the molarity of the NaOH solution? The balanced chemical equation is given in the previous problem. (a) 0.0009 M (b)0.071 M (c)0.472 M (d)0.944 M 11. If you react 1.00 L of ethane at STP with 3.00 L of oxygen at STP, how many grams of carbon dioxide will be formed? One mole of any gas occupies 22.4 L at STP. The balanced chemical equation

is... (a) 1.96 g (b)3.37 g (c)3.93 g (d)5.89 g 12. What is the percent yield of water if 0.90 g of water is obtained when 29.0 g of butane is burned in excess oxygen? The balanced chemical equation is...

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(a) 0.02% (b)2% (c)10% (d)36% 13. One of the functions of the catalytic converter in your car is to oxidize carbon monoxide to carbon dioxide. If 15.0 g of carbon monoxide reacts with 9.0 g of oxygen, how many grams of which compound remains unreacted? The balanced chemical equation is...

(a) 0.4 g of oxygen remains unreacted (b)0.8 g of carbon monoxide remains unreacted (c) 7.1 g of carbon monoxide remains unreacted (d) 8.1 g of oxygen remains unreacted

Multiple Choice Question Set 3 1. Consider the unbalanced chemical equation, F2 + H2O OF2 + HF. When the reaction is balanced with smallest integer coefficients, the coefficient for H2O is

a. 1 b.2 c.3 d.4

2. Consider the unbalanced chemical equation, Al(OH)3 + H2CO3 Al2(CO3)3 + H2O. When the reaction is balanced with smallest integer coefficients, the coefficient for H2CO3 is

a. 1 b.2 c.3 d.5

3. Consider the unbalanced chemical equation, CaCO3 + H3PO4 Ca3(PO4)2 + CO2 + H2O. When the reaction is balanced with smallest integer coefficients, the coefficient for CO2 is

a. 1 b.3 c.5 d.7

4. Consider the unbalanced chemical equation for the complete combustion of butanoic acid (C4H8O2): C4H8O2 + O2 CO2 + H2O. When the reaction is balanced using smallest integer coefficients, the coefficient for O2 is

a. 3 b.4 c.5 d.6

5. Ammonia (NH3) reacts with oxygen to produce nitrogen oxide (NO) and water. If this reaction is described with a balanced chemical equation using smallest integer coefficients, the coefficient for nitrogen oxide is

a. 1 b.2 c.3 d.4

6. Propene gas (C3H6) burns completely in oxygen to produce carbon dioxide and water as the only products. If this reaction is described with a balanced chemical equation using smallest integer coefficients, the coefficient for water is

a. 1 b.3 c.6 d.9

7. The complete combustion of liquid toluene (C7H8) in oxygen yields just carbon dioxide and water. When this reaction is balanced using smallest integer coefficients, the coefficient for water is

a. 2 b.4 c.8 d.12

8. The complete combustion of liquid propyl alcohol (C3H7OH) in oxygen yields just carbon dioxide and water. When this reaction is balanced using smallest integer coefficients, the coefficient for O2 is

a. 5 b.9 c.10 d.20

9. If the reaction, K2O2(s) + H2O(l) KOH(aq) + O2(g), is balanced using smallest integer coefficients, and these coefficients are interpreted as moles, what is the total mass of the reactants?

a. 128 g b.146 g c.238 g d.256 g

10. The complete combustion of gaseous ethane (C2H6) in oxygen yields just carbon dioxide and water. When this reaction is balanced using smallest integer coefficients, and these coefficients are interpreted as moles, what is the total mass of the reactants?

a. 62 g b.126 g c.142 g d.284 g

11. Consider the unbalanced reaction for the oxidation of sugar (sucrose) by potassium chlorate, KClO3(s) + C12H22O11(s) KCl(s) + CO2(g) + H2O(g). When this reaction is balanced, the number of moles of KClO3 needed to react completely with one mole of C12H22O11 is

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a. 2 b.4 c.8 d.12

12. When the reaction, MnO2(s) + HCl(aq) MnCl2(aq) + Cl2(g) + H2O(l), is balanced, the number of moles of Cl2 produced from 1.00 mol MnO2 and excess HCl is

a. 1 b.2 c.4 d.8

13. In the balanced chemical reaction, XeF4(g) + 2 H2O(g) Xe(g) + 4 HF(g) + O2(g), what mass of water is required to react completely with 10.0 g of XeF4?

a. 0.87 g b.1.20 g c.1.74 g d.2.40 g

14. Zinc sulfide (ZnS) reacts with oxygen to form zinc oxide (ZnO) and gaseous sulfur dioxide (SO2). What mass of SO2 is produced by the complete conversion of 10.0 g of ZnS to ZnO?

a. 3.29 g b.5.00 g c.8.35 g d.10.0 g

15. A crystalline compound of formula Mg(NO3)2�6H2O loses some of its water of hydration when it is heated, but undergoes no other chemical changes. When a 10.0 g sample is heated in this way, the residue weighs 7.19 g. What is the formula of the residue?

a. Mg2(NO3)2�11H2O b.Mg(NO3)2�4H2O

c. Mg(NO3)2�3H2O d.Mg(NO3)2�2H2O e.Mg(NO3)2�H2O

16. Potassium chlorate (KClO3) is a solid that decomposes on heating above 400°C. 7.394 g of KClO3 is subjected to prolonged heating above 400°C, after which the residue is found to weigh 4.498 g. A possible formula for the residue is

a. KCl b.KO c.KO2

d.KClO e.KClO2

17. A 3.05-g sample of an alloy of gold (Au) and copper is reacted with excess nitric acid (HNO3) to form 4.00 g of copper (II) nitrate (Cu(NO3)2). Thus the percentage of Au in the alloy, by mass, is

a. 11.9% b.33.6% c.44.4% d.55.6%

18. Assume that the following reaction between phosphorus and chlorine gas goes as far as possible to give the product, gaseous phosphorus pentachloride.P4(s) + 10 Cl2(g) 4 PCl5(g)

If 0.231 mol of P4 is reacted with an excess of Cl2, how much PCl5 will be produced?

a. 0.231 mol b.0.462 mol c.0.693 mol d.0.924 mol

19. Assume that the following reaction between phosphorus and chlorine gas goes as far as possible to give the product, gaseous phosphorus pentachloride.P4(s) + 10 Cl2(g) 4 PCl5(g). What volume of PCl5 will be produced by the reaction of 129 L of chlorine gas with excess P4, if all gas volumes are measured under the same conditions of temperature and pressure.

a. 322 L b.129 L c.51.6 L d.4.00 L

20. Assume that the following reaction between phosphorus and chlorine gas goes as far as possible to give the product, gaseous phosphorus pentachloride.P4(s) + 10 Cl2(g) 4 PCl5(g)

if 1.239 g of P4 is combined with 0.0893 mol of chlorine gas, what is the theoretical yield of PCl5?

a.0.0179 mol b.0.0357 mol c.0.0400 mol d.0.0893 mol

21. Consider the production of ammonia gas through the reaction of nitrogen with hydrogen as described by the balanced equation,N2(g) + 3 H2(g) 2 NH3(g)

How much N2 would be required to react completely with 1.50 mol of H2?

a. 14.0 g b.28.0 g c.42.0 g d.126 g

22. Consider the production of ammonia gas through the reaction of nitrogen with hydrogen as described by the balanced equation, N2(g) + 3 H2(g) 2 NH3(g)

How much NH3 would be produced by the complete reaction of 8.91 g of H2 with excess N2?

a. 25.1 g b.50.2 g c.75.3 g d.100 g

23. Consider the production of ammonia gas through the reaction of nitrogen with hydrogen as described by the balanced equation, N2(g) + 3 H2(g) 2 NH3(g)

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If all gas volumes are measured under the same conditions of pressure and temperature, what volume of ammonia gas would be produced by the complete reaction of 40.0 L of nitrogen?

a. 20.0 L b.26.7 L c.40.0 L d.80.0 L

24. Consider the reaction between hydrogen sulfide gas (H2S) and oxygen to produce sulfur dioxide gas (SO2) and water according to the balanced equation, 2 H2S(g) + 3 O2(g) 2 SO2(g) + 2 H2O(l). All gas volumes are assumed to be measured under the same fixed conditions of temperature and pressure. How much H2S will be required to react completely with 6.11 g of O2?

a. 2.71 g b.4.34 g c.6.51 g d.8.68 g

25. Consider the reaction between hydrogen sulfide gas (H2S) and oxygen to produce sulfur dioxide gas (SO2) and water according to the balanced equation, 2 H2S(g) + 3 O2(g) 2 SO2(g) + 2 H2O(l). All gas volumes are assumed to be measured under the same fixed conditions of temperature and pressure. What volume of SO2 can be produced from the complete reaction of 8.9 L of O2?

a. 2.00 L b.3.00 L c.8.9 L d.13.4 L

26. Consider the reaction between hydrogen sulfide gas (H2S) and oxygen to produce sulfur dioxide gas (SO2) and water according to the balanced equation, 2 H2S(g) + 3 O2(g) 2 SO2(g) + 2 H2O(l). All gas volumes are assumed to be measured under the same fixed conditions of temperature and pressure.If 2.50 L of H2S and 3.00 L of O2 are allowed to react completely, how much SO2 will be produced?

a. 2.00 L b.2.50 L c.2.75 L d.3.00 L

27. Consider the reaction in which carbon disulfide (CS2) is burned in oxygen to produce carbon dioxide and sulfur dioxide (SO2) according to the balanced equation, CS2(l) + 3 O2(g) CO2(g) + 2 SO2(g) What is the least amount of CS2 needed to produce 10.00 g of SO2?

a. 2.971 g b.5.943 g c.11.89 g d.23.77 g

28. Consider the reaction in which carbon disulfide (CS2) is burned in oxygen to produce carbon dioxide and sulfur dioxide (SO2) according to the balanced equation, CS2(l) + 3 O2(g) CO2(g) + 2 SO2(g) How much CO2 will be produced along with 10.00 g of SO2?

a. 3.435 g b.6.870 g c.13.74 g d.27.48 g

29. Consider the reaction in which carbon disulfide (CS2) is burned in oxygen to produce carbon dioxide and sulfur dioxide (SO2) according to the balanced equation, CS2(l) + 3 O2(g) CO2(g) + 2 SO2(g) If 10.00 g of CS2 reacts as far as possible with 15.00 g of O2, how much SO2 will be produced?

a. 8.414 g b.16.83 g c.20.02 g d.30.03 g

30. In the presence of gaseous hydrogen sulfide (H2S) and oxygen, metallic silver is converted to silver sulfide (Ag2S) according to the balanced equation,

4 Ag(s) + 2 H2S(g) + O2(g) 2 Ag2S(s) + 2 H2O(l). If 38.2 g of Ag is allowed to react with 5.60 g of H2S and 3.00 g of O2, which reactant will be the limiting reagent?

a. Ag b.H2S c.O2 d.all of these

31. The gaseous elements H2 and O2 react explosively to form water (H2O). 3.00 L of H2 is mixed with 2.00 L of O2 and the mixture is ignited in a strong steel vessel. If the gas volumes are all measured at the same temperature and pressure, which gas remains unreacted, and what is its volume?

a. O2, 0.50 L b.O2, 1.00 L c.H2, 1.00 L d.H2, 2.00 L 32. After a chemical reaction was completed, the product was carefully weighed and the mass recorded. In order to calculate the percentage yield for the reaction, what additional information is required? a. the theoretical yield of the product b.the actual yield of the product c. the molar mass of the product d.all of these 33. A series of four sequential reactions is carried out during the synthesis of an organic compound. The percentage yields for the individual reactions are listed below. Reaction Number Percentage Yield

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1 50% 2 19% 3 14% 4 16%

The overall yield for the synthesis is

a. 2.1105% b.16 % c.0.21 % d.0.0021% 34. Consider the production of acetylene gas (C2H2) by the reaction of calcium carbide (CaC2) with water as described in the balanced equation,CaC2(s) + 2 H2O(l) Ca(OH)2(s) + C2H2(g) This reaction can be assumed to go as far toward completion as possible. Also, all gas volumes refer to conditions of temperature and pressure where one mole of gas occupies 22.4 L. What volume of C2H2 will be produced by the complete reaction of 4.00 mol of CaC2?

a. 5.60 L b.22.4 L c.44.8 L d.89.6 L 35. Consider the production of acetylene gas (C2H2) by the reaction of calcium carbide (CaC2) with water as described in the balanced equation, CaC2(s) + 2 H2O(l) Ca(OH)2(s) + C2H2(g) This reaction can be assumed to go as far toward completion as possible. Also, all gas volumes refer to conditions of temperature and pressure where one mole of gas occupies 22.4 L. What mass of C2H2 will be produced by the complete reaction of 150.0 g of CaC2?

a. 2.34 g b.11.23 g c.30.42 g d.60.93 g 36. Consider the production of acetylene gas (C2H2) by the reaction of calcium carbide (CaC2) with water as described in the balanced equation,CaC2(s) + 2 H2O(l) Ca(OH)2(s) + C2H2(g) This reaction can be assumed to go as far toward completion as possible. Also, all gas volumes refer to conditions of temperature and pressure where one mole of gas occupies 22.4 L. What volume of C2H2 will be produced by the complete reaction of 28.5 g of CaC2 with 10.00 g of water?

a. 0.445 L b.0.555 L c.6.22 L d.9.97 L 37. Consider the production of acetylene gas (C2H2) by the reaction of calcium carbide (CaC2) with water as described in the balanced equation,CaC2(s) + 2 H2O(l) Ca(OH)2(s) + C2H2(g) This reaction can be assumed to go as far toward completion as possible. Also, all gas volumes refer to conditions of temperature and pressure where one mole of gas occupies 22.4 L. If 18.0 g of CaC2 reacts to produce 0.200 mol of C2H2, the percentage yield of the reaction is a. 36% b.58% c.80% d.90% 38. Consider the production of acetylene gas (C2H2) by the reaction of calcium carbide (CaC2) with water as described in the balanced equation,CaC2(s) + 2 H2O(l) Ca(OH)2(s) + C2H2(g) This reaction can be assumed to go as far toward completion as possible. Also, all gas volumes refer to conditions of temperature and pressure where one mole of gas occupies 22.4 L. If the percentage yield of C2H2 is known to be 80.0%, what volume of C2H2 is expected to be produced from 64.1 g of CaC2? a. 14.3 L b.17.9 L c.22.4 L d.28.0 L

STP GAS AND MASS STOICHIOMETRY PROBLEMS 1. How many moles of nitrogen gas is needed to react with 44.8 liters of hydrogen gas to produce

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ammonia gas? 2. How many liters of ammonia are produced when 89.6 liters of hydrogen are used in the above reaction? 3. Ten grams of calcium carbonate was produced when carbon dioxide was added to lime water (calcium hydroxide in solution). What volume of carbon dioxide at STP was needed? 4. When 11.2 liters of hydrogen gas is made by adding zinc to sulfuric acid, what mass of zinc is needed? 5. What volume of ammonia at STP is needed to add to water to produce 11 moles of ammonia water? 6. How many grams of carbonic acid is produced when 55 liters of carbon dioxide is pressed into water? 7. magnesium hydroxide + ammonium sulfate magnesium sulfate + water + ammonia How much (grams) magnesium hydroxide do you need to use in the above reaction to produce 500 liters of ammonia? 8. How much strontium bromide is needed to add to chlorine gas to produce 75 liters of bromine? 9. What mass of ammonium chlorate is needed to decompose to give off 200 liters of oxygen? 10. Your car burns mostly octane, C8H18, as a fuel. How many liters of oxygen is needed to burn a kilogram of octane? 11. copper + sulfuric acid copper II sulfate + water + sulfur dioxide How many moles of copper are needed to produce 1000 L of SO2? 12. What volume of oxygen is needed to burn a pound of magnesium? 13. How many grams of sodium do you have to put into water to make 30 liters of hydrogen at STP? 14. ammonia gas and hydrogen chloride gas combine to make ammonium chloride. What volume of ammonia at STP is needed to react with 47.7 liters of hydrogen chloride at STP? 15. How many liters of oxygen are needed to burn 10 liters of acetylene?

ANSWERS TO GAS AND MASS STOICHIOMETRY PROBLEMS 1. 0.667 mol 2. 59.7 L 3. 2.24 L 4. 32.7 g

5. 246 L 6. 152 g 7. 651 g 8. 828 g

9. 604 g 10. 2.46 kL 11. 44.6 mol 12. 210 L

13. 61.6 g 14. 47.7 L 15. 25 L

PROBLEMS ON CONCENTRATION AND DENSITY 1. The lead brick on my desk measures 3 by 5 by 11 cm. Lead has a density of 11.34 g/cc. How many lead atoms are in that block? 2. The lab technician at the Planter's Peanut packing factory takes a bag of peanuts, puts water into it to dissolve the salt, and dilutes the solution to one liter. She then takes ten ml of that solution and titrates it against 0.132 M silver nitrate. One bag sample takes 31.5 ml of silver nitrate to endpoint. What mass of salt was in the bag? 3. What is the concentration of sugar (C12H22O11) if twenty grams are dissolved in enough water to make 2 liters? 4. Methyl alcohol (CH3OH) has a density of 0.793 kg/l. What volume of it is needed to add to water to make five liters of 0.25 M solution? 5. Magnesium has a density of 1.741 g/cc. What volume of Mg will burn in 20 liters of oxygen at 2.1 atm and 0°C? 6. Uranium metal can be purified from uranium hexafluoride by adding calcium metal. Calcium metal has a density of 1.54 g/cc. Uranium has a density of 18.7 g/cc. What mass of uranium do you get for a Kg of Ca? What volume of uranium do you get for a cubic meter of calcium? 7. What volume of 0.27 M sodium hydroxide is needed to react with 29.5 ml of 0.55 M phosphoric acid? 8. What volume of carbon dioxide is produced at 1 atm and 87 °C when 1.6 liters of methyl alcohol burns? What volume of liquid water is produced in this reaction? 9. Seven kilograms of mercury II oxide decomposes into mercury and oxygen. Mercury has a density of 13.6 g/cc/ What volume of mercury is produced? 10. Water and calcium oxide produce calcium hydroxide. How many grams of calcium hydroxide are made if you add 275 liters of water to enough calcium oxide? 11. Gasoline (C7H16) has a density of 0.685 kg/liter. How many liters of oxygen at 37 °C and 950 mmHg are needed to burn 15 liters of gasoline? 12. Sodium hydroxide and hydrochloric acid combine to make table salt and water. 14 mL of 0.1 M sodium hydroxide is added to an excess of acid. How many moles of table salt are made? How many grams of salt is that? 13. 50 mL of 0.25 M copper II sulfate evaporates to leave CuSO4·5H2O. (That is the pentahydrate crystal of copper II sulfate.) What is the mass of this beautiful blue crystal from the solution? 14. Chlorine gas is bubbled into 100 mL of 0.25 M potassium bromide solution. This produces potassium chloride and bromine gas. The bromine (which dissolves in water) is taken from the solution and

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measured at 27 °C and 825 mmHg. What is the volume of bromine? 15. 95.0 mL of 0.55 M sulfuric acid is put on an excess of zinc. This produces zinc sulfate and hydrogen. How many grams of zinc sulfate are made? 16. 27.6 mL of a 0.190 M solution of silver nitrate and 15.4 mL of an unknown (but excess) amount of sodium chloride combine to make a white precipitate silver chloride and some dissolved sodium nitrate. (a) How many moles of silver chloride are made? (b) How many grams of silver chloride is that? (c) How many moles of sodium nitrate are made? (d) What is the concentration of sodium nitrate in the final solution? 17. How many grams of potassium permanganate, KMnO4, is needed to make 1.72 liters of 0.29 M solution? 18. By my calculations, a drop of ethyl alcohol, C2H5OH , in an olympic-sized swimming pool produces a 1.20 E-10 M solution of alcohol in water. A drop is a twentieth of a mL. How many molecules of ethyl alcohol are in a drop of the water in the pool? 19. 93.0 mL of 0.150 M magnesium hydroxide is added to 57.0 mL of 0.4 M nitric acid. (Magnesium nitrate and water are formed. What is the concentration of the magnesium nitrate after the reaction?

ANSWERS TO PROBLEMS ON CONCENTRATION AND DENSITY 1. 5.44 E24 atoms 2. 24.3 g 3. 0.0292 M 4. 0.0504 L

5. 52.3 ml(cc) Mg 6a. 1.98 kg of U 6b. 1.63 E6 mL 7. 180 mL

8a. 1.17 kL CO2 8b. 1.43 L 9. 0.477 L 10. 1.13 E 6 g

11. 23.0 kL 12a. 1.4 E-3 mols 12b. 0.0819 g 13. 3.12 g

14. 284 mL 15. 8.44 g 16a. 5.24E-3 mol 16b. 0.752 g

16c. 5.24E-3 mols 16d. 122 mmolar 17. 78.8 mg 18. 3.61E9 molecules

19. 0.152 M

PROBLEMS USING COMPLETE ROADMAP 1. How many liters of ammonia at 0 °C and 25 atm. are produced when 10 g of hydrogen is combined with nitrogen? 2. How many milliliters of hydrogen at 0 deg C and 1400 mmHg are made if magnesium reacts with 15 mL of 6 M sulfuric acid? 3. How many atoms are in 25 liters of fluorine gas at 2.85 atm and 450 °C? 4. Liquid butane (C4H10 has a density of 0.60 g/cc. It burns to make carbon dioxide at 120 °C. What volume of carbon dioxide is produced at one atm when 350 liters of liquid butane burns? 5. Isopropyl alcohol, C3H7OH , makes a good fuel for cars. What volume of oxygen at 785 mmHg and 23 °C is needed to burn 8.54 E25 molecules of isopropyl alcohol? 6. How many moles of NaCl are in a liter of a 0.15 M NaCl solution? (0.15 M NaCl is physiological saline when sterilized.) 7. How many grams of NaCl must you put into a 50 liter container to make a physiological saline solution? 8. Chlorine gas is bubbled into 100 mL of 0.25 M potassium bromide solution. This produces potassium chloride and bromine gas. The bromine dissolves completely in the water. What is the concentration of bromine? 9. 95 mL of 0.55 M sulfuric acid is put on an excess of zinc. This produces zinc sulfate and hydrogen. How many grams of zinc sulfate are made? 10. Methyl alcohol (CH3OH) has a density of 0.793 Kg/L. What volume of it is needed to add to water to make twenty-five liters of 0.15 M solution? 11. Magnesium has a density of 1.741 g/cc. What volume of Mg will burn to produce a kilogram of magnesium oxide? 12. What volume of water vapor is produced at 716 mmHg and 87°C when 2.6 liters of methyl alcohol burns?

ANSWERS TO PROBLEMS USING COMPLETE ROADMAP 1. 2.99 L 2. 1.10 E3 mL 3. 1.45 E24 atoms 4. 4.67 E5 L

5. 1.50 E4 L 6. 0.15 moles 7. 439 g 8. 0.125 M

9. 8.44 g 10. 151 mL 11. 0.346 L 12. 1.29 E5 L

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