Basic Natural Sciences III..

Solutions, stoichiometry, calculations. Redox reactions.

Example 1. (CLM/36/14.)How many cm3 0.35 mass fraction phosphoric acid must be diluted to250 cm3 to get a 50 g/dm3 solution? The density of the original solutionis 1.216 g/cm3.

Dilution: water is added to the original solution.

Original solution Prepared solution

V2 = 250 cm3

= 50 g/dm3

V1 = ? cm3

w1 = 0.35 = 1.216 g/cm3

=msolute

Vsolution

·Vsolution =

= 50 · 0.25 =

=12.5 g

1) msolute =

2) msolute = 12.5 gw =

msolute

msolution

msolute3) msolution =w =

12.50.35 = 35.71 g

= mV

msolution4) Vsolution =

=

= 29.37 cm335.711.216

=

Calculations and reaction equations

3

aA + bB cC + dD

m, M, V, ,, m/m% etc.

is given

m, M, V, ,, m/m% etc.is the question

?

nA

step-1: write the correct reaction equation!step-2: balance the reaction equation!step-3: calculate the amount of the well-known substance!step-4: calculate the amount of the substance that we want to know!step-5: answer the question (m, M, V, r, g, m/m% etc)!

nD =nA

a ·d

Example 2. (CLM/48/6.)How many cm3 0.15 g/cm3 mass concentration of sodium thiosulfate solutioncan discolourate the 80 cm3 0.15 g/cm3 mass-concentration of I2 (in Lugolsolution). Reaction equation:

2Na2S2O3 + I2 = Na2S4O6 + 2NaI

2Na2S2O3 + I2 = Na2S4O6 + 2NaI

V2 = 80 cm3

= 0.15 g/cm3V1 = ? cm3

= 0.15 g/cm3

We need n to use thereaction

equation!

4.72·10-2 mol

nI2 = mI2 / MI2 = 12 / 254 = 4.72·10-2 mol2)4) mNa2S2O3 = nNa2S2O3 ·MNa2S2O3 =

= 9.45·10-2 · 158 = 14.93 g

2 · 4.72·10-2 mol3)= 9.45·10-2 mol

mI2 = V2 · 2 = 80 · 0.15 = 12 g1)

VNa2S2O3 = mNa2S2O3 / 1 = 14.93 / 0.15 = 99.5 cm35)

Molar volume

The molar volume (Vm) is the volume occupied by one mole of a substance (chemical element or chemical compound) at a given temperature and pressure.

Avogadro's law: "equal volumes of all gases, at the same temperature and pressure, have the same number of molecules".

Under standard conditions (25 °C, 0.1 Mpa)

Vm = 24.5 dm3/mol

The volume of 1 mol gas under st. conditions is always 24.5 dm3.

V = n · Vm n = Vm

VorReminder

m = n · M

n = Mm

Example 3. (CLM/49/13.) How many grams of iron sulfide is needed to reactwith hydrochloric acid to produce 3.5 dm3 standard dihydrogen sulfide gas?

FeS + 2HCl FeCl2 + H2S0.143 mol0.143 mol

VH2S = 3.5 dm3 (USC)

nH2S = = = 0.143 molVH2S,

24.5

3.5

24.5

mFeS = nFeS * MFeS = 0.143*88 = 12.584 g

MFeS = 88 g / mol

m = n·M

Example 4. (CLM/48/4.) We bubble 3.5 dm3 standard hydrogen chloride and hydrogen fluoride gas mixture into silver nitrate solution. 10.0 grams of precipitate will form. What volume % was the gas mixture?

First step: write the reaction equation(s)!

HCl + AgNO3 AgCl + HNO3

(The AgF is not precipatate, it means the 10 g precipitate whichformed in the reaction is the AgCl.)

Second step: calculate the amount of the precipitate (AgCl).

mAgCl = 10 g n = mM (MAgCl = 143.4 g/mol)

(To be continued on next page )

nAgCl = 10/143.4 = 0.07 mol

HCl + AgNO3 AgCl + HNO30.07 mol0.07 mol

(1 mol AgCl forms from 1 mol HCl; now 0.07 mol AgCl formed, so the gascontained the same amount of HCl.)

(V/V)%HCL = * 100 = * 100 = 49%VHCl

VTOTAL

1.7153,5

(V/V)%HF = 100-49 = 51%

Example 5. (CLM/56/4.) We bubble 3.5 dm3 standard hydrogen chloride and hydrogen fluoride gas mixture into silver nitrate solution. 10.0 grams of precipitate will form. What volume % was the gas mixture?

nHCl = 0.07 molThis gas was in standard state! V = n·Vm

VHCL = nHCl·Vm= 0.07·24.5 = 1.715 dm3

Vm = 24.5 dm3/mol

Example 6. (CLM/64/7.) How many cm3 of 10 mass% sodium hydroxide solution( = 1.11 g/cm3) must be added to 60 cm3 25 g/dm3 mass concentration lead nitratesolution to make the initially formed precipitate dissolve completely?

Reaction equation:

Pb(NO3)2 + 4NaOH Na2[Pb(OH)4] + 2NaNO3

Vsolution = 60 cm3 = 0.06 dm3

Pb(NO3)2 = 25 g/dm3 =mPb(NO3)2

Vsolution

mPb(NO3)2 = ·Vsolution = 25·0.06 = 1.5 g

n =mM

nPb(NO3)2 = mPb(NO3)2/MPb(NO3)2 = 1.5/331 = 0.0045 mol

MPb(NO3)2 = 331 g / mol

0.0045 mol 4·0.0045 == 0.018 mol

Pb(NO3)2 + 4NaOH Na2[Pb(OH)4] + 2NaNO3

0.0045 mol 4·0.0045 == 0.018 mol

nNaOH = 0.018 mol

MNaOH = 40 g / molmNaOH = nNaOH·MNaOH = 0.018·40 = 0.72 g (solute)

m = n·M

m/m% = ·100mNaOH

msolutionmsolution = ·100 = ·100 = 7.2 g

mNaOH

m/m%0.7210 (solution)

Example 6. (CLM/64/7.) How many cm3 of 10 mass% sodium hydroxide solution( = 1.11 g/cm3) must be added to 60 cm3 25 g/dm3 mass concentration lead nitratesolution to make the initially formed precipitate dissolve completely?

= mV

V = =msolution

7.21.11

= 6.5 cm3

Example 7. (CLM/49/11.) How many cm3 0.2 g/cm3 mass concentrationhydrogen peroxide solution is needed to oxidize 9.5 g lead-sulfide to lead-sulfate?

Reaction equation:

PbS + 4 H2O2 → PbSO4 + 4 H2O

At first we will calculate the amount of the PbS:

mPbS = 9.5 gMPbS = 239.2 g/mol nPbS = = = 0.04 moln =

mM

mPbS

MPbS

9.5239.2

0.04 mol 0.16 mol(We see in the reaction 1 mol PbS reactswith 4 mol H2O2. We have 0.04 mol PbS, so we need 4·0.04 = 0.16 mol H2O2. )

MH2O2 = 34 g/molnH2O2 = 0,16 mol mH2O2 = MH2O2·nH2O2 = 34·0.16 = 5.44 g

m = n·M

mH2O2 = 5.44 gH2O2 = 0.2 g/cm3

m H2O2

H2O2VH2O2 = = 5.44 / 0.2 = 27.2 cm3

=mH2O2

Vsolution

Example 7. (CLM/49/11.) How many cm3 0.2 g/cm3 mass concentrationhydrogen peroxide solution is needed to oxidize 9.5 g lead-sulfide to lead-sulfate?

Reaction equation:

PbS + 4 H2O2 → PbSO4 + 4 H2O

Oxidizing and reducing agents

Hydrazine N2H4

Sulfite compounds SO32-

Carbon C

Hydrogen H2

Hydride compounds H-

Metals Zn, Fe, Na

Oxidizing agents Reducing agents

Oxygen O2

Halogens F2, Cl2, Br2, I2Nitric acid HNO3

Potassiumpermanganate KMnO4

Hydrogen peroxide H2O2

but: KMnO4 + H2O2 + H2SO4 K2SO4 + MnSO4 + O2 + H2O !!!!

Potassium permanganate in acid solution changes to Mn2+ (MnSO4)Potassium permanganate in neutral or alkaline solution changes to MnO2

Oxidation numbers

4. In molecules the algebraic sum of the oxidation numbers of atoms equals zero. In case of ions the sum of oxidation numbers equals the electrical charge of the ion.

Rules to determine the oxidation numbers:

1. The oxidation number of atoms in elements is zero.

2. The oxidation number of alkali metals (column I. in periodic system) is always +1 (except elemental form).

The oxidation number of alkali earth metals (column II. in periodic system) is always +2 (except elemental form).

The oxidation number of aluminium is +3 (except elemental).

3. The oxidation number of hydrogen usually +1 (except elemental form)and hydrides, when it is -1 (e.g. LiH, CaH2, etc.).

The oxidation number of oxygen usually is -2 (except elemental form)and peroxides, when it is -1 (e.g. H2O2).

Oxidation numbers - examples

+1 ( )·3-2 2·(+1)+x+3·(-2) = 0

2+x-6 = 0x-4 = 0x = +4

2 ·( )+4

H2SO3

+2

(OCl)2charge: -2

OClcharge: -1

OCl--2 +1

Ca(OCl)2 Ca(OCl)2

+2 -2 +1

2 ·( )+1 ( )·7-2+6 2·(+1)+2·x+7·(-2) = 0

2+2x-14 = 02x-12 = 02x = 12

x = 6

K2Cr2O7

SO42-

Cu2+ Cu2++2

SO42-( )·4-2

x+4·(-2) = -2x-8 = -2

x = 6

+6

CuSO4+2 -2+6CuSO4

Fe + HNO3 = Fe(NO3)3 + NO + H2O

1. step: calculate the oxidation numbers of the atoms

Fe + HNO3 = Fe(NO3)3 + NO + H2O0 +1 +5-2 +3 +5 -2 +2 -2 -2+1

Balance the next reaction equation:

Fe + HNO3 = Fe(NO3)3 + NO + H2O

2. step: search which oxidation numbers are changed

+3 -3

0 +1 +5-2 +3 +5 -2 +2 -2 -2+1

3. step: check the number of the atoms whose oxidation state changed

(In this case there is just 1 nitrogen atom in one HNO3,therefore we multiply with one.)

1·(+3) 1·(-3)

Fe + HNO3 = Fe(NO3)3 + NO + H2O0 +1 +5-2 +3 +5 -2 +2 -2 -2+1

1Fe + 1HNO3 = Fe(NO3)3 + NO + H2Obecause 3 : 3 = 1 : 1 (it’s simplier)

5. step: write the coefficients of the products whose oxidation number changed

1Fe + 1HNO3 = 1Fe(NO3)3 + 1NO + H2O6. step: in this case there are 1*3 = 3 „extra” nitrogen in the right side (in theFe(NO3)3, whose oxidation number is not changed in the reaction. We must add this number to the left side.

1Fe + 1+3HNO3 = 1Fe(NO3)3 + 1NO + H2O7. step: balance everything else

1Fe + 4HNO3 = 1Fe(NO3)3 + 1NO + 2H2O

4. step: write the coefficients of the reactants whose oxidation number changed

1·(+3) = 3 1·(-3) = -3

3Fe + 3HNO3 = Fe(NO3)3 + NO + H2O

Balance the next reaction equation:

NH3 + Cl2 = N2 + NH4Cl

NH3 + Cl2 = N2 + NH4Cl-3 +1 0 0 +1 -1-3

1·3 2·1

2NH3 + 3Cl2 = 1N2 + 6NH4Cl

8NH3 + 3Cl2 = 1N2 + 6NH4Cl

2+6

Balance the next reaction equation:

Al + NaOH + H2O = Na[Al(OH)4] + H2

-2Al + NaOH + H2O = Na[Al(OH)4] + H2

+3+1 +1 +1+1-2 -20 0+1

1·3 1·1 OH H H+ + OH-

H2

[Na[Al(OH)4]

1Al + 1NaOH + 3H2O = 1Na[Al(OH)4] + 1.5H2

Because fraction is not elegant, we multiply everything with 2:

2Al + 2NaOH + 6H2O = 2Na[Al(OH)4] + 3H2

The end

20

CLM/37/22. What is the mole fraction of a 3 mol/dm3 concentration and 1.073 g/cm3 density sodium chloride solution?

We can say Vsolution = 1 dm3 = 1000 cm3

x = ?

c = 3 mol/dm3

= 1.073 g/cm3

If n, m or V is NOT given in the task, we canchoose a starting value of one of them.

c =nV

= mV

x =nsolute

nsolution

NaCl

nNaCl = c·V = 3·1=3 mol

nsolute + nsolvent

nsolute=

nsolution = ?

msolution = ·V = 1.073·1000 = 1073 g

nsolution = nsolute + nsolvent

3 mol ? mol

CLM/37/22. What is the mole fraction of a 3 mol/dm3 concentration and 1.073 g/cm3 density sodium chloride solution?

msolution1073 g =

nNaCl = 3 mol

mNaCl = 175.5 g

mH2O = 1073 – 175.5 = 897.5 g

n =mM

mNaCl = nNaCl·MNaCl = 3·58.5 = 175.5 g

MNaCl = 58.5 g/mol

n =mM nH2O =

mH2O

MH2O

MH2O = 18 g/mol

897.518

= = 49.86 mol

CLM/37/22. What is the mole fraction of a 3 mol/dm3 concentration and 1.073 g/cm3 density sodium chloride solution?

nsolution = nsolute + nsolvent = 3 + 49.86 = 52.86 mol

3 mol 49.86 mol

x =nsolute

nsolution nsolute + nsolvent

nsolute=

352.86

= = 0.0567

Balance the next reaction equations:

KMnO4 + KI + H2O = KOH + MnO2 + I2

MnSO4 + NaOH + KNO3 = Na2MnO4 + Na2SO4 + KNO2 + H2O

(Correct answer on the next pages)

Homework

Balance the next reaction equation:

KMnO4 + KI + H2O = KOH + MnO2 + I2-2+1 0+7 +1-1 -2+1 -2+1 +1 +4 +2

1·11·3

1KMnO4 + 3KI + 2H2O = 4KOH + 1MnO2 + 1.5I2

2KMnO4 + 6KI + 4H2O = 8KOH + 2MnO2 + 3I2

Because fraction is not elegant, we multiply everything with 2:

KMnO4 + KI + H2O = KOH + MnO2 + I2

Solution:

Balance the next reaction equation:

MnSO4 + NaOH + KNO3 = Na2MnO4 + Na2SO4 + KNO2 + H2O-2+2 +6

1·4 1·2

-2+1 +1 +5+1 -2 +6+1 -2 +6+1 -2 +3+1 -2 -2+1

Solution:

MnSO4 + NaOH + KNO3 = Na2MnO4 + Na2SO4 + KNO2 + H2O

2MnSO4+8NaOH+4KNO3 = 2Na2MnO4+2Na2SO4+4KNO2+4H2O