Quantities in Chemical Reactions STOICHIOMETRY PROBLEMS Section 5.4

What is Stoichiometry? The study of quantitative relationships within chemical reactions A balanced equation is the key to stoichiometry! Tools you’ll need for this chapter: Writing proper formulas and balanced equations Finding molar mass Converting from mass to moles and vice versa

Balanced Reaction Equations and the Mole Ratio Consider this balanced reaction equation: 4 Al(s) + 3 O2(g) → 2 Al2O3(s) The coefficients of this reaction represent: - the number of reacting PARTICLES or - the number of reacting MOLES These numbers are FIXED, the ratio of reacting substances NEVER changes.

Stoichiometry Problems There are three types of stoichiometry problems we will encounter: Mole-Mole problems (1 conversion factor) Mass-Mole problems (2 conversion factors) Mass-Mass problems (3 conversion factors) given required

Mole-Mole Problems Step 1: Write a BALANCED EQUATION Step 2: Determine the mole ratio from the coefficients in the equation. Mole ratio = moles of required substance moles of given substance Step 3: Multiply the amount of moles of the given substance by the mole ratio

Mole-Mole Problems 2 H2O 2 H2 + O2 Example: How many moles of water can be formed from 0.5 mol H2? 2 mol H2O 0.5 mol H2 0.5 mol H2O x = 2 mol H2

Mole-Mole Practice 3 2 CuSO4 Al Al2(SO4)3 Cu + + How many moles of copper(II) sulfate will react with 0.5 moles of aluminum? Mole ratio 0.5 mol Al 3 mol CuSO4 2 mol Al = x 0.8 mol CuSO4

Mass-Mole Problems 2 H2O 2 H2 + O2 Example: How many moles of water can be formed from 48.0 g O2?

Setting up the given information 2 H2 + O2 2 H2O Mass 48.0g Mole Ratio Moles ?

Mass-Mole Problems Step 1: Write a BALANCED EQUATION. Step 2: Convert the mass of your given substance to moles using molar mass. Step 3: Determine the moles of your required substance using the mole ratio.

Setting up the given information 2 H2 + O2 2 H2O Step 1 Mass 48.0g Mole Ratio Moles ? Step 2: M = 32.00g/mol Step 3

Mass-Mole Problems 2 H2O 2 H2 + O2 Example: How many moles of water can be formed from 48.0 g O2? 1 mol O2 2 mol H2O 48.0 g O2 3.00 mol H2O x x = 1 mol O2 32.00 g O2

Mass-Mole Practice CuSO4 Al Al2(SO4)3 Cu 3 2 How many moles of aluminum sulphate can be produced from 13.5 g of aluminum? CuSO4 Al Al2(SO4)3 Cu 3 2 + Mole ratio 1 mol Al 1 mol Al2(SO4)3 13.5 g Al x x 0.250 mol Al2(SO4)3 = 26.98 g Al 2 mol Al

Mass-Mole Practice 3 2 Ca AlCl3 3 CaCl2 Al + + How many moles of calcium chloride will be produced if 5.7g of calcium is used up in the reaction?

Mass-Mole Practice 3 2 Ca AlCl3 3 CaCl2 Al + + How many moles of calcium chloride will be produced if 5.7g of calcium is used up in the reaction? 1 mol Ca 3 mol CaCl2 5.7 g Ca x x 0.14 mol CaCl2 = 40.08 g Ca 3 mol Ca

Mass-Mass Problems 2 H2O 2 H2 + O2 Example: How many grams of water can be formed from 48.0 g O2?

Setting up the given information 2 H2 + O2 2 H2O Step 1 Mass 48.0g Mole Ratio Moles ? Step 2: M = 32.00g/mol Step 4: M = 18.02g/mol Step 3

Mass-Mass Problems 2 H2O 2 H2 + O2 Example: How many grams of water can be formed from 48.0 g O2? 1 mol O2 2 mol H2O 18.02 g H2O 48.0 g O2 54.1 g H2O x x x = 32.00 g O2 1 mol O2 1 mol H2O

Mass-Mass Practice 3 2 Ca AlCl3 3 CaCl2 Al + + How much aluminum is produced (in grams) when 1.9g of calcium reacts with aluminum chloride?

Mass-Mass Practice 3 2 Ca AlCl3 3 CaCl2 Al + + How much aluminum is produced (in grams) when 1.9g of calcium reacts with aluminum chloride? 1.9 g Ca 2 mol Al 3 mol Ca = x 0.85 g Al 1 mol Ca 40.08 g Ca 26.98 g Al 1 mol Al