Chapter 9Chapter 9

StoichiometryStoichiometry

The calculation of quantities The calculation of quantities in chemical reactionsin chemical reactions

ParticlesParticles

NN2(g)2(g) + 3H + 3H2(g)2(g) 2NH 2NH3(g)3(g)

1 molecule of N1 molecule of N2 2 reacts with 3 reacts with 3 molecules of Hmolecules of H22 to produce 2 to produce 2 molecules of NHmolecules of NH33

Ratio : 1:3:2 particlesRatio : 1:3:2 particles

MolesMoles

1 mole = 6.02 x 101 mole = 6.02 x 102323 representative particlesrepresentative particles

Based on the reaction – mole ratio Based on the reaction – mole ratio is 1:3:2is 1:3:2

* Coefficients are the relative # * Coefficients are the relative # of molesof moles

MassMass Law of conservation of Law of conservation of

massmass

1:3 21:3 2

28g + 6 g 34 g28g + 6 g 34 g

N = 2x14 H = 2x3 NHN = 2x14 H = 2x3 NH33 = 17x2 = 17x2

VolumeVolume

Assume STP Assume STP 22.4 L/mol22.4 L/mol

ExampleExample

2H2H22S + 3OS + 3O22 2SO 2SO22 + + 2H2H22OO

Mass of ReactantsMass of Reactants

2 mol x 34.1 g = 68.2g2 mol x 34.1 g = 68.2g

molmol

3 mol x 32 g = 96g3 mol x 32 g = 96g

mol mol

164.2 g164.2 g

Mass of ProductsMass of Products

2 mol x 64.1g = 128.2 g2 mol x 64.1g = 128.2 g

molmol

2 mol x 18g = 36 g2 mol x 18g = 36 g

molmol

164.2 g164.2 g

Volume of gases at STP Volume of gases at STP

ReactantsReactants

2 mol H2 mol H22S x 22.4 L = 44.8 L HS x 22.4 L = 44.8 L H22SS

1 mol1 mol

3 mol O3 mol O22 x 22.4 L = 67.2 L O x 22.4 L = 67.2 L O22

1 mol1 mol

ProductsProducts

2 mol SO2 mol SO22 x 22.4 L = 44.8 L SO x 22.4 L = 44.8 L SO22

1 mol1 mol

2 mol H2 mol H22O x 22.4 L = 44.8 L HO x 22.4 L = 44.8 L H22OO

1 mol1 mol

Mole to Mole CalculationsMole to Mole Calculations

How many moles of ammonia How many moles of ammonia are produced when 0.06 mol of are produced when 0.06 mol of nitrogen reacts with hydrogen?nitrogen reacts with hydrogen?

NN22 + 3H + 3H22 2NH 2NH33

0.06 mol N0.06 mol N22 x 2mol NH x 2mol NH3 3

1 mol N1 mol N22

= .12mol NH= .12mol NH33

ExampleExample

Aluminum Oxide is formed fromAluminum Oxide is formed from

aluminum and oxygenaluminum and oxygen

a) Write a balanced equationa) Write a balanced equation

4Al + 3O4Al + 3O22 2Al 2Al22OO33

2.3 mol Al2.3 mol Al22OO33 x 4 mol Al x 4 mol Al

2 mol Al2 mol Al22OO33

= 4.6mol Al= 4.6mol Al

b) How many moles of Al are b) How many moles of Al are needed to form 2.3 mol of Alneeded to form 2.3 mol of Al22OO33 ??

.84mol Al x 3mol O.84mol Al x 3mol O22

4mol Al4mol Al

c) How many moles of Oxygen are c) How many moles of Oxygen are required to react completely with required to react completely with 0.84 mol of Al ?0.84 mol of Al ?

= .63 mol O= .63 mol O22

17.2 mol O17.2 mol O22 x 2mol Al x 2mol Al22OO33

3mol O3mol O22

= 11.47 mol Al= 11.47 mol Al22OO33

d) Calculate the # of moles of Ald) Calculate the # of moles of Al22OO33 formed when 17.2 mol of Oformed when 17.2 mol of O22 reacts reacts with Al ?with Al ?

Mass – Mass CalculationsMass – Mass Calculations

- GFM GFM - Ex) HEx) H22

1 mol or 2g1 mol or 2g

2g 1 mol2g 1 mol

ExampleExample

Calculate the number of grams Calculate the number of grams of ammonia produced by the of ammonia produced by the reaction of 5.40 g of hydrogen reaction of 5.40 g of hydrogen with nitrogen.with nitrogen.

NN2(g)2(g) + 3H + 3H2(g)2(g) 2NH 2NH3(g)3(g)

5.40g H5.40g H22 x 1mol H x 1mol H22 x 2mol NH x 2mol NH33 x 17g NH x 17g NH33

2g H2g H22 3mol H 3mol H22 1mol NH 1mol NH33

= 30.6g NH= 30.6g NH33

These problems are solved the same way These problems are solved the same way as mole – mole problemsas mole – mole problems

StepsSteps

1)1) Change mass to moles G Change mass to moles G (given)(given)

2)2) Change moles of G to moles of Change moles of G to moles of WW

3)3) Change moles of W to grams of Change moles of W to grams of WW

ExampleExample

2C2C22HH22 + 5O + 5O22 4CO 4CO22 + 2H + 2H22OO

a) How many grams of Oa) How many grams of O22 are required to burn 13g are required to burn 13g of Cof C22HH2 2 ??

= 40.0g = 40.0g OO22

13g C13g C22HH22 x 1mol C x 1mol C22HH22 x 4mol CO x 4mol CO22 x 44g CO x 44g CO22

26g C26g C22HH22 2mol C 2mol C22HH2 2 1mol CO1mol CO22

= 44g CO= 44g CO22

b) How many grams of COb) How many grams of CO22 and H and H22O are O are produced when 13g of Cproduced when 13g of C22HH22 react with react with the oxygen. the oxygen.

ContinuedContinued

13g C13g C22HH22 x 1mol C x 1mol C22HH22 x 2mol H x 2mol H22O x 18g O x 18g HH22OO

26g C26g C22HH22 2mol C 2mol C22HH22 1mol 1mol HH22OO

= 9g H= 9g H22OO

Other Stoichiometric Other Stoichiometric CalculationsCalculations

1)1) Mass – VolumeMass – Volume

2)2) Volume – VolumeVolume – Volume

3)3) Particle – MassParticle – Mass

ExampleExample

Tin plus hydroflouric acid yields Tin(II)flouride Tin plus hydroflouric acid yields Tin(II)flouride and hydrogen gasand hydrogen gas

SnSn(s)(s) + 2HF + 2HF(g)(g) SnF SnF2(s)2(s) + H + H2(g)2(g)

966g SnF966g SnF22

a) How many grams of SnFa) How many grams of SnF22 can be can bemade by reacting 7.42 x 10made by reacting 7.42 x 102424 molecules moleculesof HF with tin?of HF with tin?

4.42 L H4.42 L H22

b) How many L of Hb) How many L of H22 (STP) are produced (STP) are produced by reacting 23.4g of Sn with HF?by reacting 23.4g of Sn with HF?

28.4 L HF28.4 L HF

c) How many L of HF are needed to produce 14.2 c) How many L of HF are needed to produce 14.2 L of HL of H22(STP) ?(STP) ?

1.08 x 101.08 x 102424 molecules H molecules H22

d) How many molecules of Hd) How many molecules of H22 are produced are produced by the reaction of Sn with 80 L of HF (STP) ?by the reaction of Sn with 80 L of HF (STP) ?

Limiting ReagentLimiting Reagent

ExampleExample 1lb of swiss & 1lb of pastrami 1lb of swiss & 1lb of pastrami

and 14 slices of bread and 14 slices of bread

Limiting reagent - breadLimiting reagent - bread

•Excess reagent – Excess reagent – luncheon meatsluncheon meats

•You can only make 7 You can only make 7 sandwichessandwiches

ExampleExample Sodium Chloride is prepared by the Sodium Chloride is prepared by the

reaction of sodium metal with reaction of sodium metal with chlorine gaschlorine gas

2Na2Na(s)(s) + Cl + Cl2(g)2(g) 2NaCl 2NaCl(s)(s)

What will occur when 6.70mol of Na What will occur when 6.70mol of Na reacts with 3.20 mol of Clreacts with 3.20 mol of Cl2 2 ??

a) What is the limiting reagent ?a) What is the limiting reagent ?

ClCl22 is the limiting is the limiting reagentreagent.

= 3.35 mol Cl= 3.35 mol Cl22 required required

* 3.35 mol Cl* 3.35 mol Cl22 are required to are required to completely react with 6.70mol Na but completely react with 6.70mol Na but you only have 3.20 mol Clyou only have 3.20 mol Cl2.2.

6.70mol Na x 1mol Cl6.70mol Na x 1mol Cl22

2mol Na2mol Na

= 6.40 mol NaCl= 6.40 mol NaCl

b) How many moles of NaCl are produced?b) How many moles of NaCl are produced?

3.20mol Cl3.20mol Cl22 x 2mol NaCl x 2mol NaCl

1mol Cl1mol Cl22

.30mol Na is excess.30mol Na is excess

= 6.40mol Na= 6.40mol Na

c) How much of the excess reagent c) How much of the excess reagent remains un-reacted ?remains un-reacted ?

3.20mol Cl3.20mol Cl22 x 2mol x 2mol Na Na 1mol 1mol ClCl22

6.70mol Na 6.70mol Na

- 6.40mol Na- 6.40mol Na

ExampleExample CC22HH44 + 3O + 3O22 2CO 2CO22 + +

2H2H22OO

If 2.70 mol CIf 2.70 mol C22HH44 reacts with 6.30 reacts with 6.30 mol 0mol 022

a) What is the limiting reagent?a) What is the limiting reagent?

the limiting reagent is Othe limiting reagent is O22

2.7mol C2.7mol C22HH44 x 3mol O x 3mol O2 2 = 8.1mol O = 8.1mol O22

1mol C1mol C22HH44

* You need 8.1mol O* You need 8.1mol O22 to react with 2.7 mol to react with 2.7 mol CC22HH44

= 4.2mol H= 4.2mol H22OO

b) Calculate moles of Hb) Calculate moles of H22O producedO produced

6.3 mol O6.3 mol O2 2 x 2mol H x 2mol H22OO

3mol O3mol O22

= 2.1mol C= 2.1mol C22HH44

c) Calculate moles of excess reagentc) Calculate moles of excess reagent

6.30mol O6.30mol O22 x 1mol C x 1mol C22HH44

3mol O3mol O22

0.60mol C0.60mol C22HH44

Final StepFinal Step

2.7mol C2.7mol C22HH44

- 2.1mol C- 2.1mol C22HH44

ExampleExample

a) How many grams of Ha) How many grams of H22 can be can be produced when 4g HCl is added produced when 4g HCl is added to 3g of Mg?to 3g of Mg?

=0.11g H=0.11g H22

4g HCl x 1mol HCl = 0.11mol HCl4g HCl x 1mol HCl = 0.11mol HCl 36g HCl36g HCl

3g Mg x 1mol Mg = .12mol Mg3g Mg x 1mol Mg = .12mol Mg

24.3g Mg24.3g Mg

.11mol HCl x 1mol H.11mol HCl x 1mol H2 2 x 2g x 2g HH22

2mol HCl 1mol 2mol HCl 1mol HH2 2

= 1.23 L H= 1.23 L H22

b) Assuming STP, what is the b) Assuming STP, what is the volume of Hvolume of H22??

0.11g H0.11g H22 x 1mol x 22.4L x 1mol x 22.4L

2g 1mol2g 1mol

Percent YieldPercent Yield

Theoretical YieldTheoretical Yield – the maximum – the maximum amount of product that could be amount of product that could be formed from a given amount of formed from a given amount of reactantreactant

Actual YieldActual Yield – Amount of product – Amount of product that forms when the reaction is that forms when the reaction is carried out in the lab.carried out in the lab.

Percent YieldPercent Yield – Ratio of actual – Ratio of actual yield to theoretical yieldyield to theoretical yield

% Yield = actual yield x % Yield = actual yield x 100100

theoretical yieldtheoretical yield

ExampleExample

CaCOCaCO33 CaO + CO CaO + CO22

*Calculate the % yield of CaO if *Calculate the % yield of CaO if 24.89g CaCO24.89g CaCO33 is heated to give is heated to give 13.1g of CaO13.1g of CaO

= 13.9g CaO= 13.9g CaO

24.8g CaCO24.8g CaCO33 x 1mol CaCO x 1mol CaCO3 3 x x

100.1g CaCO100.1g CaCO33

1mol CaO x 56.1g CaO1mol CaO x 56.1g CaO

1mol CaCO1mol CaCO33 1mol CaO 1mol CaO

94.2%94.2%

Percent YieldPercent Yield

13.113.1 13.913.9

xx 100 =100 =

ExampleExample

What is the % yield of copper if 3.74g What is the % yield of copper if 3.74g of copper is produced when 1.87g of of copper is produced when 1.87g of Aluminum is reacted with an excess Aluminum is reacted with an excess of copper(II)sulfate?of copper(II)sulfate?

2Al + 3CuSO2Al + 3CuSO44 Al Al22(SO(SO44))33 + + 3Cu3Cu

*3.74g Cu*3.74g Cu

1.87g Al x 1mol Al x 3mol Cu x 1.87g Al x 1mol Al x 3mol Cu x 63.54g63.54g

26.98g Al 2mol Al 1mol Cu26.98g Al 2mol Al 1mol Cu

= 6.61g Cu= 6.61g Cu

Percent YieldPercent Yield

3.74g Cu x 100 = 3.74g Cu x 100 =

6.61g Cu6.61g Cu56.6 56.6 %%