Chapter 9

Stoichiometry The calculation of quantities in chemical reactions

Particles N2(g) + 3H2(g) 2NH3(g) 1 molecule of N2 reacts with 3 molecules of H2 to produce 2 molecules of NH3 Ratio : 1:3:2 particles

Moles 1 mole = 6.02 x 1023 representative particles Based on the reaction – mole ratio is 1:3:2 * Coefficients are the relative # of moles

Mass Law of conservation of mass 1:3 2 28g + 6 g 34 g 1:3 2 28g + 6 g 34 g N = 2x14 H = 2x3 NH3 = 17x2

Volume Assume STP 22.4 L/mol

Example 2H2S + 3O2 2SO2 + 2H2O

Mass of Reactants 2 mol x 34.1 g = 68.2g mol 3 mol x 32 g = 96g

Mass of Products 2 mol x 64.1g = 128.2 g mol 2 mol x 18g = 36 g

Volume of gases at STP Reactants 2 mol H2S x 22.4 L = 44.8 L H2S 1 mol 3 mol O2 x 22.4 L = 67.2 L O2

Products 2 mol SO2 x 22.4 L = 44.8 L SO2 1 mol 2 mol H2O x 22.4 L = 44.8 L H2O

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

N2 + 3H2 2NH3 0.06 mol N2 x 2mol NH3 1 mol N2 = .12mol NH3

Example Aluminum Oxide is formed from aluminum and oxygen a) Write a balanced equation 4Al + 3O2 2Al2O3

b) How many moles of Al are needed to form 2.3 mol of Al2O3 ? 2.3 mol Al2O3 x 4 mol Al 2 mol Al2O3 = 4.6mol Al

c) How many moles of Oxygen are required to react completely with 0 c) How many moles of Oxygen are required to react completely with 0.84 mol of Al ? .84mol Al x 3mol O2 4mol Al = .63 mol O2

d) Calculate the # of moles of Al2O3 formed when 17 d) Calculate the # of moles of Al2O3 formed when 17.2 mol of O2 reacts with Al ? 17.2 mol O2 x 2mol Al2O3 3mol O2 = 11.47 mol Al2O3

Mass – Mass Calculations GFM Ex) H2 1 mol or 2g 2g 1 mol

Example Calculate the number of grams of ammonia produced by the reaction of 5.40 g of hydrogen with nitrogen. N2(g) + 3H2(g) 2NH3(g)

5.40g H2 x 1mol H2 x 2mol NH3 x 17g NH3 2g H2 3mol H2 1mol NH3 = 30.6g NH3 These problems are solved the same way as mole – mole problems

Steps Change mass to moles G (given) Change moles of G to moles of W Change moles of W to grams of W

Example a) How many grams of O2 are required to burn 13g of C2H2 ? 2C2H2 + 5O2 4CO2 + 2H2O a) How many grams of O2 are required to burn 13g of C2H2 ? = 40.0g O2

b) How many grams of CO2 and H2O are produced when 13g of C2H2 react with the oxygen. 13g C2H2 x 1mol C2H2 x 4mol CO2 x 44g CO2 26g C2H2 2mol C2H2 1mol CO2 = 44g CO2

Continued 13g C2H2 x 1mol C2H2 x 2mol H2O x 18g H2O 26g C2H2 2mol C2H2 1mol H2O = 9g H2O

Other Stoichiometric Calculations Mass – Volume Volume – Volume Particle – Mass

Example Tin plus hydroflouric acid yields Tin(II)flouride and hydrogen gas Sn(s) + 2HF(g) SnF2(s) + H2(g)

a) How many grams of SnF2 can be made by reacting 7.42 x 1024 molecules of HF with tin? 966g SnF2

b) How many L of H2 (STP) are produced by reacting 23.4g of Sn with HF? 4.42 L H2

c) How many L of HF are needed to produce 14.2 L of H2(STP) ? 28.4 L HF

d) How many molecules of H2 are produced by the reaction of Sn with 80 L of HF (STP) ? 1.08 x 1024 molecules H2

Limiting Reagent Example 1lb of swiss & 1lb of pastrami and 14 slices of bread You can only make 7 sandwiches Limiting reagent - bread Excess reagent – luncheon meats

Example Sodium Chloride is prepared by the reaction of sodium metal with chlorine gas 2Na(s) + Cl2(g) 2NaCl(s) What will occur when 6.70mol of Na reacts with 3.20 mol of Cl2 ?

a) What is the limiting reagent ? 6.70mol Na x 1mol Cl2 2mol Na = 3.35 mol Cl2 required * 3.35 mol Cl2 are required to completely react with 6.70mol Na but you only have 3.20 mol Cl2. Cl2 is the limiting reagent.

b) How many moles of NaCl are produced? 3.20mol Cl2 x 2mol NaCl 1mol Cl2 = 6.40 mol NaCl

3.20mol Cl2 x 2mol Na 1mol Cl2 = 6.40mol Na 6.70mol Na - 6.40mol Na c) How much of the excess reagent remains un-reacted ? 3.20mol Cl2 x 2mol Na 1mol Cl2 = 6.40mol Na 6.70mol Na - 6.40mol Na .30mol Na is excess

Example C2H4 + 3O2 2CO2 + 2H2O If 2.70 mol C2H4 reacts with 6.30 mol 02 a) What is the limiting reagent?

the limiting reagent is O2 2.7mol C2H4 x 3mol O2 = 8.1mol O2 1mol C2H4 * You need 8.1mol O2 to react with 2.7 mol C2H4 the limiting reagent is O2

6.3 mol O2 x 2mol H2O 3mol O2 = 4.2mol H2O b) Calculate moles of H2O produced 6.3 mol O2 x 2mol H2O 3mol O2 = 4.2mol H2O

c) Calculate moles of excess reagent 6.30mol O2 x 1mol C2H4 3mol O2 = 2.1mol C2H4

Final Step 2.7mol C2H4 - 2.1mol C2H4 0.60mol C2H4

Example a) How many grams of H2 can be produced when 4g HCl is added to 3g of Mg?

3g Mg x 1mol Mg = .12mol Mg 24.3g Mg .11mol HCl x 1mol H2 x 2g H2 4g HCl x 1mol HCl = 0.11mol HCl 36g HCl 3g Mg x 1mol Mg = .12mol Mg 24.3g Mg .11mol HCl x 1mol H2 x 2g H2 2mol HCl 1mol H2 =0.11g H2

b) Assuming STP, what is the volume of H2? 0.11g H2 x 1mol x 22.4L 2g 1mol = 1.23 L H2

Percent Yield Theoretical Yield – the maximum amount of product that could be formed from a given amount of reactant

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

Percent Yield – Ratio of actual yield to theoretical yield % Yield = actual yield x 100 theoretical yield

Example CaCO3 CaO + CO2 *Calculate the % yield of CaO if 24.89g CaCO3 is heated to give 13.1g of CaO

24.8g CaCO3 x 1mol CaCO3 x 100.1g CaCO3 1mol CaO x 56.1g CaO 1mol CaCO3 1mol CaO = 13.9g CaO

Percent Yield 13.1 13.9 94.2% x 100 =

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

2Al + 3CuSO4 Al2(SO4)3 + 3Cu 1.87g Al x 1mol Al x 3mol Cu x 63.54g *3.74g Cu 1.87g Al x 1mol Al x 3mol Cu x 63.54g 26.98g Al 2mol Al 1mol Cu = 6.61g Cu

Percent Yield 3.74g Cu x 100 = 6.61g Cu 56.6 %