Section 2: Stoichiometric Calculations

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Section 2: Stoichiometric Calculations A car’s air bags are powered by the rapid reaction of a small mass of sodium azide (Na3N) into a large volume of N2 gas. (all in less than a second) You will use a balanced chemical equation to calculate the amount of reactant needed or product formed. Print 1-8,10-13

Stoichiometric Calculations In a typical stoichiometric problem, 1) Convert grams of given to moles of given 2) Use the mol ratio to convert moles of given to moles of wanted 3) Convert moles of wanted to grams of wanted Note – you can start or stop a stoichiometry problem anywhere on the above path, depending on the question.

Mg + 2 HCl → MgCl2 + H2 mol ratio Moles to Moles How many moles of HCl are required to produce 5.5 mol MgCl2? Mg + 2 HCl → MgCl2 + H2 2 mol HCl 1 mol MgCl2 5.5 mol MgCl2 x = 11 mol HCl mol ratio 3

Mg + 2 HCl → MgCl2 + H2 mol ratio molar mass Moles to Grams How many grams of H2 are produced from 4.0 mol HCl? Mg + 2 HCl → MgCl2 + H2 1 mol H2 2 mol HCl 2.02 g H2 1 mol H2 4.0 mol HCl x x = 4.04 g H2 mol ratio molar mass 4

Mg + 2 HCl → MgCl2 + H2 molar mass mol ratio Grams to Moles How many mol of H2 are produced from 48 g Mg? Mg + 2 HCl → MgCl2 + H2 1 mol Mg 24.31 g Mg 1 mol H2 1 mol Mg 48 g Mg x x = 1.97 mol H2 molar mass mol ratio 5

Mg + 2 HCl → MgCl2 + H2 mol ratio molar mass molar mass Grams to Grams How many grams of H2 are produced from 18 g HCl? Mg + 2 HCl → MgCl2 + H2 1 mol HCl 36.46 g HCl 1 mol H2 2 mol HCl 2.02 g H2 1 mol H2 18 g HCl x x x = 0.499 g H2 mol ratio molar mass molar mass 6

Have we learned it yet? Try these on your own - 4 NH3 + 5 O2  6 H2O + 4 NO a) How many moles of H2O can be made using 1.6 mol NH3? b) what mass of NH3 is needed to make 0.75 mol NO? c) how many grams of NO can be made from 47 g of NH3?

Answers 4 NH3 + 5 O2  6 H2O + 4 NO 2.4 mol H2O 13 g NH3 83 g NO b) c) 6 mol H2O 4 mol NH3 x 2.4 mol H2O = 1.6 mol NH3 4 mol NH3 4 mol NO x 17.04 g NH3 1 mol NH3 x 13 g NH3 = 0.75 mol NO 1 mol NH3 17.04 g NH3 x 4 mol NO 4 mol NH3 x 30.01 g NO 1 mol NO x 47 g NH3 83 g NO =