Stoichiometry “Cookbook Chemistry”

There are many similarities between baking and chemistry Starting point Raw ingredients Reactants Ending Point Delicious dish Products How much Recipe Mole Ratio

S’more Stoichiometry

C3H8(g) + O2(g)  CO2(g) + H2O(g) The combustion of propane (C3H8(g)): C3H8(g) + O2(g)  CO2(g) + H2O(g) Balance the equation to determine the mole ratio. Mole ratio =

The synthesis of ammonia (NH3(g)): N2(g) + H2(g)  NH3(g) Mole ratio = If 2 mol of N2 react in an excess of H2, how many moles of NH3 will be produced?

CH3OH(l) + O2(g)  CO2(g) + H2O(g) The combustion of methanol (CH3OH(l)): CH3OH(l) + O2(g)  CO2(g) + H2O(g) Mole ratio = If 3.5 mol of methanol are present, How many moles of O2 are reacted? How many moles of H2O are produced?

Stoichiometry Races! CH4 + 2 O2 → CO2 + 2 H2O How many moles of CO2 will be produced from 2 mol of CH4 and 4 mol of O2 ? How many moles of H2O will be produced from 1 mol of CH4 and 18 mol of O2 ?

Calculating Unknown Masses We already know how to convert from mole to mass Now we can use our balanced equation to calculate masses of reactants and products We use a stoichiometry table

 72.0 g of O2 will be produced from 1.50 mol KClO3 What mass of O2 will be produced through the decomposition of 1.50 mol KClO3? 2 KClO3  2 KCl + 3 O2 MM 32.00 g/mol n 1.50 mol = 3 molO2 X 1.50 molKClO3 2 molKClO3 = 2.25 molO2 m = 2.25 mol X 32.00 g 1 mol = 72.0 g  72.0 g of O2 will be produced from 1.50 mol KClO3

What mass of Cu2S will be produced if 2.00 g of Cu are reacted?  8 Cu2S(s) MM 63.55 g/mol 159.17 g/mol n = 2.00 g X 1 mol 63.55 g = 0.03147 mol = 8 molCu2S X 0.03147 molCu 16 molCu = 0.01574 mol m 2.00 g = 0.01574 mol X 159.17g 1 mol = 2.50 g  2.50 g of Cu2S will be produced from 2.00 g Cu

Practice Makes Perfect  Worksheet A Worksheet B