Solving a Stoichiometry Problem Balance the equation. Convert given to moles. Determine which reactant is limiting. Use moles of limiting reactant and mole ratios to find moles of desired product. Convert from moles to grams, molecules or Liters.

Practice Test 2 (Ch. 9) 2 Al + 6 HCl  2 AlCl3 + 3 H2 6 2 1. How many moles of hydrochloric acid (HCl) would be produced from 4.18 moles of aluminum chloride (AlCl3)? mol HCl 4.18 mol Use Mole Ratio (Balanced Equation) mol HCl 6 4.18 mol AlCl3 X = mol HCl 12.54 2 mol AlCl3

Practice Test 2 (Ch. 9) 2 Al + 6 HCl  2 AlCl3 + 3 H2 2 6 2. If you wanted to completely react 7.5 moles of hydrochloric acid (HCl), how many moles of Aluminum would be needed? 7.5 mol mol Al Use Mole Ratio (Balanced Equation) mol Al 2 7.5 mol HCl X = mol Al 2.5 6 mol HCl

H Cl Al 2 Al + HCl  AlCl3 + H2 6 2 3 Given: Want: 3. How many grams of hydrochloric acid (HCl) would be needed to completely react 76.2 g of aluminum foil? Given: Want: 1 mol Al 6 mol HCl 36.46 g HCl 76.2 g Al = 309 g HCl X X X 26.98 g Al 2 mol Al 1 mol HCl Convert to moles (molar mass = 1 mole) Mole ratio (Balanced Eqn.) Convert to grams (1 mole = molar mass) 1.01 x 1 HCl 35.45 +35.45 = 36.46 g 1 H Hydrogen 1.01 17 Cl Chlorine 35.45 13 Al Aluminum 26.98 Do the Calculation: 76.2 ÷ 26.98 x 6 ÷ 2 x 36.46 = 308.9234989 = 309

H Cl 2 Al + HCl  AlCl3 + H2 6 2 3 Given: Want: 4. How many grams of hydrochloric acid (HCl) would be needed to produce 15.0 dm3 of hydrogen gas? Given: Want: 1 mol H2 6 mol HCl 36.46 g HCl 15.0 dm3 H2 = 48.8 X g HCl X X 22.4 dm3 H2 3 mol H2 1 mol HCl Convert to moles (22.4 dm3 = 1 mole) Mole ratio (Balanced Eqn.) Convert to grams (1 mole = molar mass) 1.01 x 1 HCl 35.45 +35.45 = 36.46 g 1 H Hydrogen 1.01 17 Cl Chlorine 35.45 Do the Calculation: 15.0 ÷ 22.4 x 6 ÷ 3 x 36.46 = 48.83035714 = 48.8

C C(s) + SO2(g)  CS2(l) + CO(g) 5 2 1 4 Given: Want: 5. How many molecules of carbon disulfide (CS2) would be produced by completely reacting 8.00 grams of carbon (C) with excess sulfur dioxide gas (SO2)? Given: Want: 1 mol C 1 mol CS2 6.02 x 1023 molecules CS2 8.00 g C = X X X 8.02 x 1022 molecules CS2 12.01 g C 5 mol C 1 mol CS2 6 C Carbon 12.01 Convert to moles (molar mass = 1 mole) Mole ratio (Balanced Eqn.) Convert to molecules (1 mole = 6.02 x 1023) Do the Calculation: 8.00 ÷ 12.01 ÷ 5 x 6.02 x 1023 = 8.01998 x 1022

C(s) + SO2(g)  CS2(l) + CO(g) 5 2 4 6. What volume (in Liters) of sulfur dioxide gas (SO2) is needed to produce 6.0 L of carbon monoxide gas (CO)? Given: Want: 1 mol CO mol SO2 22.4 L SO2 2 = 3.0 L SO2 6.0 L CO X X X 22.4 L CO 4 mol CO 1 mol SO2 Convert to moles (22.4 L = 1 mole) Mole ratio (Balanced Eqn.) Convert to Liters (1 mole = 22.4 L) Do the Calculation: 6.0 ÷ 22.4 x 2 ÷ 4 x 22.4 = 3

N 1 N2(g) + H2(g)  NH3(g) 3 2 Given: Want: 7. What volume (in Liters) of ammonia gas (NH3) would be produced from 70.05 grams of nitrogen (N2) at STP Given: Want: 1 mol N2 2 mol NH3 22.4 L NH3 = 112.0 L NH3 70.05 g N2 X X X 28.02 g N2 1 mol N2 1 mol NH3 7 N Nitrogen 14.01 Convert to moles (molar mass = 1 mole) Mole ratio (Balanced Eqn.) Convert to Liters (1 mole = 22.4 L) Do the Calculation: 70.05 ÷ 28.02 x 2 x 22.4 = 112

Solving a Stoichiometry Problem Compare the two reactants from balance equation to moles given for each reactant in the problem. Solving a Stoichiometry Problem Reactants 2 C4H10 (g) + 13 O2 (g)  8 CO2 (g) + 10 H2O (l) If 2.5 moles of butane is mixed with 18.0 moles of O2, and the mixture is ignited, determine: a. The limiting reactant Ideal situation (From Bal.Eqn.) Moles given in Problem “I have” “I need” C4H10 2.5 0.14 C4H10 O2 = 0.15 2 = = = O2 1 18 13 1 0.14 is less than 0.15 so you don’t have enough butane which means 2.5 moles of butane, C4H10 is the limiting reagent.

Solving a Stoichiometry Problem Compare the two reactants from balance equation to moles given for each reactant in the problem. Solving a Stoichiometry Problem Reactants 2 C4H10 (g) + 13 O2 (g)  8 CO2 (g) + 10 H2O (l) If 2.5 moles of butane is mixed with 18.0 moles of O2, and the mixture is ignited, determine: b. The excess reactant Ideal situation (From Bal.Eqn.) Moles given in Problem “I have” “I need” O2 18 7.2 O2 C4H10 = 6.5 13 = = = C4H10 1 2.5 2 1 7.2 is more than 6.5 so you have “extra” moles of O2 making it the excess reagent.

Solving a Stoichiometry Problem Limiting 2 C4H10 (g) + 13 O2 (g)  8 CO2 (g) + 10 H2O (l) 2 8 If 2.5 moles of butane is mixed with 18.0 moles of O2, and the mixture is ignited, determine: If 2.5 moles of butane is mixed with 18.0 moles of O2, and the mixture is ignited, determine: c. The moles of carbon dioxide (CO2) produced. c. The moles of carbon dioxide (CO2) produced. Identify what you want to find Start with the limiting reagent mol CO2 2.5 mol C4H10 x 10. = ________ mol CO2 mol C4H10 Use Mole Ratio (Bal. Equation)

Solving a Stoichiometry Problem Limiting Excess 2 C4H10 (g) + 13 O2 (g)  8 CO2 (g) + 10 H2O (l) 2 13 If 2.5 moles of butane is mixed with 18.0 moles of O2, and the mixture is ignited, determine: If 2.5 moles of butane is mixed with 18.0 moles of O2, and the mixture is ignited, determine: d. How many grams of excess reactant left over ? d. How many grams of excess reactant left over? 1st find out how much excess gets used up. Identify what you want to find Start with the limiting reagent mol O2 2.5 mol C4H10 x 16.25 = ________ mol O2 mol C4H10 Amount Used up Use Mole Ratio (Bal. Equation)

Solving a Stoichiometry Problem Limiting Excess 2 C4H10 (g) + 13 O2 (g)  8 CO2 (g) + 10 H2O (l)C If 2.5 moles of butane is mixed with 18.0 moles of O2, and the mixture is ignited, determine: 18.0 mol O2 d. How many grams of excess reactant left over ? d. How many grams of excess reactant left over ? - mol O2 16.25 Amount Used up = 1.75 mol O2 Excess Reagent (original amount) Excess moles Left over

Solving a Stoichiometry Problem Limiting Excess 2 C4H10 (g) + 13 O2 (g)  8 CO2 (g) + 10 H2O (l) If 2.5 moles of butane is mixed with 18.0 moles of O2, and the mixture is ignited, determine: d. How many grams of excess reactant left over ? d. How many grams of excess reactant left over? 1 mole = Molar mass (Periodic table) 3 sig.figs. 32 g O2 1.75 mol O2 x 56.0 56 = ________ g O2 1 mol O2 grams of excess reactant. Use molar mass (Periodic table)

Solving a Stoichiometry Problem Limiting 2 2 C4H10 (g) + 13 O2 (g)  8 CO2 (g) + 10 H2O (l) 10 If 2.5 moles of butane is mixed with 18.0 moles of O2, and the mixture is ignited, determine: If 2.5 moles of butane is mixed with 18.0 moles of O2, and the mixture is ignited, determine: e. How many grams of water will theoretically be produced fro this reaction? e. How many grams of water will theoretically be produced fro this reaction? Identify what you want to find Start with the limiting reagent 1 mole = molar mass (Periodic Table) mol H2O 18.02 g H2O 2.5 mol C4H10 x x = ________ g H2O 1 mol H2O mol C4H10 Use Mole Ratio (Bal. Equation) 225.25 Theoretical Yield

Solving a Stoichiometry Problem Limiting Excess 2 C4H10 (g) + 13 O2 (g)  8 CO2 (g) + 10 H2O (l) If 2.5 moles of butane is mixed with 18.0 moles of O2, and the mixture is ignited, determine: f. Determine the % yield if the actual amount of water produced during this reaction is 91.10 g. 91.10 g Actual Yield = % Yield = X 100 X 100 Theoretical Yield 225.25 g from previous Problem = 40.4 %