Chapter 10

•Calculate the molar mass of the following compounds:a) NaOH - 40g/molb) AuCl3 – 303.32g/mol

c) (NH4)2SO4 – 132.16g/mol

Q1

•Convert 3.4 moles of NH4 into grams of NH4.

Q2

=

3.4 moleof NH4 61g

18.04gof NH4

1 mol of NH4

Convert 4.2 grams of Ca(NO3)2 into moles of Ca(NO3)2.

Q3

=

4.2g ofCa(NO3)2 0.026mol

1molof Ca(NO3)2

164.1g of Ca(NO3)2

Convert 7.8 L of N2 into moles.

Q4

=

7.8L ofN2 0.35 mol

1molof N2

22.4L of N2

Convert 3.5 mol of H2 into liters

Q5

=

3.5 mol ofH2 78 L of H2

22.4Lof H2

1 mol ofH2

Convert 1.4 moles of CO2 into molecules of CO2.

Q6

=

1.4 mol ofCO2 8.43 x 1023

6.02 x 1023

moleculesof CO2

1 mol ofCO2

Convert 2.3 X 1025 atoms of Mg(NO3)2 into grams of Mg(NO3)2.

Q7

=

2.3 x 1025

moleculesof Mg(NO3)2

5.7 X 103g

1 mol of Mg(NO3)2

6.02 x 1023

moleculesof Mg(NO3)2

148.31g of Mg(NO3)2

1 mol of Mg(NO3)2

Q8

What is the percent composition of sulfur in Al2(SO4)3?

= % Comp Part (3 x molar mass of S)

Whole (molar mass of Al2(SO4)3)

X 100%

96.18g

342.15gX 100% = 28.1% S

What is the percent composition of carbon in pentane, C5H12?

Q9

= % Comp Part (5 x molar mass of C)

Whole (molar mass of C5H12)

X 100%

60.05g

72.15gX 100% = 83.23% C

q10

Chapter 12

(Ch 12) Q1Mole-Mole Problem:

According to the following equation, how many moles of Al react with 5.2 moles of H2SO4?

2Al + 3H2SO4 Al2(SO4)3 + 3H2

=

5.2 molesof H2SO4 3.5 mole

2 mol of Al

3 molesof H2SO4

(Ch 12) Q2Mole-Mole Problem:

According to the following equation, how many moles of oxygen are needed to form 3.7 mol Al2O3?

4Al + 3O2 2Al2O3

=

3.7 molesof Al2O3 5.6 mole

3 mol of O2

2 molesof Al2O3

(Ch 12) Q3Mass-Mole Problem:

According to the equation in problem #2, how many moles of Al2O3 are formed when 7.8 grams of aluminum completely reacts with oxygen?

4Al + 3O2 2Al2O3

=7.8g of Al

0.14mole1 mol of Al

26.98g of Al4 mol ofAl

2 mol of Al2O3

(Ch 12) Q4Mass-Mole-Mole-Mass Problem:

According to the following equation, how many grams of CO2 would be produced if 45g of C2H2 completely reacted with oxygen?

2C2H2 + 5O2 4CO2 + 2H2O

=

45g ofC2H2 152.1g CO2

1 mol C2H2

26.0g ofC2H2

2 mol C2H2

4 mol CO2

1 mol CO2

44.0g of CO2

(Ch 12) Q5a/b/c

In the equation below 3.4g of Al reacts with 4.2g of O2 to form aluminum oxide.a.) What is the limiting reactant?

4Al + 3O2 2Al2O3

=

3.4g ofAl 6.4g Al2O3

1 mol Al

27.0g ofAl

4 mol Al

2 mol Al2O3

1 mol Al2O3

101.9g Al2O3

=

4.2g ofO2 8.9g Al2O3

1 mol O2

32.0g ofO2

3 mol O2

2 mol Al2O3

1 mol Al2O3

101.9g Al2O3

TheoreticalYield

Limiting Reagent

ExcessReagent

(Ch 12) Q5d

d.) How much of the excess reactant is left over?

4Al + 3O2 2Al2O3

=

6.4g ofAl2O3 3.02g O2

4 mol Al

3 mol O2

1 mol O2

32.0g O2

Excess Reactant (O2) = 4.2-3.02g = 1.18g

101.9g Al2O3

1 mol Al2O3

(Ch 12) Q5e

e) If the actual yield of aluminum oxide is 1.24g Al2O3 what is the percent yield?

4Al + 3O2 2Al2O3

= % Yield Actual yield (grams)

Theoretical yield (grams)X 100%

1.24g of Al2O3

6.4g of Al2O3

X 100% = 19.3%

(Ch 12) Q6

Sodium chloride decomposes through electrolysis by the below. What is the percent yield of sodium produced if 45g of NaCl reacts and produces an actual yield of 6.2g of Na?

2NaCl 2Na + Cl2

=

45g ofNaCl 17.67g of Na

1 mol NaCl

58.44g ofNaCl

2 mol NaCl

2 mol Na

1 mol Na

23.0g Na

Theoretical Yield

(Ch 12) Q6

Sodium chloride decomposes through electrolysis by the below. What is the percent yield of sodium produced if 45g of NaCl reacts and produces an actual yield of 6.2g of Na?

2NaCl 2Na + Cl2

6.2g of Na (actual yield)

17.67g Na (Theoretical yield)

X 100% = 35%

(Ch 12) Q7The combustion of propane, C3H8 is shown in the equation below. Identify the limiting reactant and the volume of CO2 formed when 6 L of C3H8 reacts with 12 L of O2 to produce CO2 gas and H2O vapor at STP.C3H8(g) + 5O2(g) 3CO2(g) + 4H2O (g)

=

6L of C3H8 18L of CO2

3L of CO2

1L of C3H8

=

12L of O2 7.2L of CO2

3L of CO2

5L of O2

Limiting Reactant

(Ch 12) Q8From problem #7, how much of excess reactant is left unreacted?

C3H8(g) + 5O2(g) 3CO2(g) + 4H2O (g)

Excess Reactant (C3H8) = 6-2.4 = 3.6L

= 2.4L of C3H8 needed

1L of C3H8

3L of CO2

7.2L of CO2

(Ch 12) Q9

From problem #7, if the actual yield was 6.9 L of CO2, what is the percent yield?

C3H8(g) + 5O2(g) 3CO2(g) + 4H2O (g)

6.9L of CO2 (actual yield)

7.2L CO2 (Theoretical yield)

X 100% = 95.8%

(Ch 12) Q10a

In the reaction below, 5.50g of Mg reacts with 3.25g of O2 to produce MgO.a) Balance this equation:

2 Mg + 1 O2 2 MgO

(Ch 12) Q10b/c/d

In the reaction below, 5.50g of Mg reacts with 3.25g of O2 to produce MgO.b) Which is the theoretical yield of MgO?

2 Mg + 1 O2 2 MgO

=

5.5g ofMg 9.12g MgO

1 mol Mg

24.3g ofMg

2 mol Mg

2 mol MgO

1 mol MgO

40.3g MgO

=

3.25g ofO2 8.19g MgO

1 mol O2

32.0g ofO2

1 mol O2

2 mol MgO

1 mol MgO

40.3g MgO

Limiting Reactant

ExcessReactant

(Ch 12) Q10e

In the reaction below, 5.50g of Mg reacts with 3.25g of O2 to produce MgO.e) How much of the excess reactant is left unreacted?

2 Mg + 1 O2 2 MgO

=8.19g MgO

4.94g Mg1 mol MgO

40.3g MgO 2 mol MgO

2 mol Mg

1 mol Mg

24.3g Mg

Excess Reactant (Mg) = 5.50-4.94= 0.56g

(Ch 12) Q10f

In the reaction below, 5.50g of Mg reacts with 3.25g of O2 to produce MgO.f) Calculate the percent yield of MgO if the actual yield was 6.80g MgO.

2 Mg + 1 O2 2 MgO

6.8g of MgO (actual yield)

8.19g MgO (Theoretical yield)

X 100% = 83.0%

Chapter 14

Ch14 Q5 P1V1 = P2V2

A gas has a volume of 50cm3 at a pressure of 200 mm Hg. If the pressure is changed to 190 mm Hg, what is the new volume of that sample of gas?

P2

P1V1

P1V1 = P2V2 V2=P1V1

P2

=200 x 50

190

V2 = 53cm3

Ch14 WP Q6V1/T1 = V2/T2

If 400mL of nitrogen is collected at 30oC, what will the new volume be if the temperature is increased to 50oC?

T2

T1V1

V1 V2=V1T2

T1

=400 x 323

303

V1 = 426mL

T1 =

V2 T2

Ch14 Q7P1/T1 = P2/T2

A cylinder of carbon dioxide left outside has a pressure of 3.5 atm and a temperature of 50oC. If the cylinder is taken indoors and cooled to 25oC, what will the new pressure be?

T2

P1 P2=P1T2

T1=

3.5 x 298

323

P2 = 3.2 atm

T1 =

P2 T2

P1 T1

Ch14 Q8 P1V1 = P2V2

A gas has a volume of 125 mL at a pressure of 2 atm. If the pressure is changed to 7.2 atm, what is the new volume of that sample of gas?

P2

P1V1

P1V1 = P2V2 V2=P1V1

P2

=2 x 125

7.2

V2 = 35mL

Ch14 Q9P1/T1 = P2/T2

To what temperature must a sample of oxygen gas at 350 K and pressure of 4.2 atm be cooled to so the final pressure is 1.25 atm?

P2

P1 T2 =P2T1

P1

=1.25 x 350

4.2

T2 = 104K

T1 =

P2 T2

P1

T1

Ch14 Q10V1/T1 = V2/T2

A gas has a volume of 20 L at a temperature of 350 K. What will the volume of the gas occupy if the Kelvin temperature is increased to 423 K?

T2

T1V1

V1 V2=V1T2

T1

=20 x 423

350

V2 = 24 L

T1 =

V2 T2

Chapter 16

Ch16 Q1M = mol/L

Calculate the molarity of a 3.4 mol of Na2CO3 dissolved in a 100mL solution?

3.4 mole

0.100L= 34M

Ch16 Q2M = mol/L

Calculate the molarity of a solution prepared by dissolving 14.5 g of BaCl2 in 250 mL of solution.

=

14.5g ofBaCl2 0.069mole of BaCl2

1 mol BaCl2

208.2g of BaCl2

0.069 mole

0.250L= 0.028M

Ch16 Q3ML = mol

Calculate the number of grams of solute needed to make a 150 mL solution of 0.40M H2SO4.

=0.4M x 0.150L 0.06mole of H2SO4

0.06mole x 89.08g/mole= 5.88g of H2SO4

Ch16 Q4L = mol/M

If you have 5.2 g of CaCl2 and you want to make a 2.5 M solution, what volume in liter should CaCl2 be diluted to?

=

5.2g ofCaCl2 0.047mole of CaCl2

1 mol CaCl2

110.98g of CaCl20.047 mole

2.5M= 0.019L

Ch16 Q5M1V1 = M2V2

You want to make a solution with a final concentration of 2.25M HCl from a stock solution of 6.0M HCl. How many milliliters of the stock solution are required to make a solution with a final volume of 50mL?

M1

M2

V2

M1V1 = M2V2

M2V2 V1 =M1

=2.25 x 50

6.0

V1 = 18.75mL

Ch16 Q6M1V1 = M2V2

You want to make a 2.50 L solution of 3.50 M HNO3 from a 9.0M HNO3 solution. How many mL of the 9.0M HNO3 should you use? (1L = 1,000mL)

M1

v2M2

M1V1 = M2V2

M2V2 V1 =M1

=3.50 x 2.50

9.0

V1 = 972mL

Ch16 Q7M1V1 = M2V2

You want to make a 250 mL solution of 0.75 M H2SO4 from a 6.0M H2SO4 solution. How many mL of the 6.0M H2SO4 should you use?

M1

v2M2

M1V1 = M2V2

M2V2 V1 =M1

=0.75 x 250

6.0

V1 = 31.25mL

Ch16 Q8M1V1 = M2V2

How many mL of a 15.8M HNO3 stock solution are required to make a 3.0M HNO3 solution with a final volume of 250mL? M2

M1

V2

M1V1 = M2V2

M2V2 V1 =M1

=3.0 x 250

15.8

V1 = 47.5mL