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Vocabulary Terms Stoichiometry: Limiting Reagent: Excess Reagent:
the calculation of quantities in chemical equation. Limiting Reagent: the reactant that determines the amount of product in a reaction. Excess Reagent: A quantity of a reactant that is more than enough to react with a limiting reagent.

Percent Yield: Actual Yield: Theoretical Yield:
the ratio of the actual yield to the theoretical yield. Actual Yield: the amount of product formed when a reaction is carried out in the lab. Theoretical Yield: the calculated amount of product formed during a reaction.

Important Things To Remember:
Always start with a balanced equation! Remember that you have to go through moles to convert from substance A to substance B.

1. _1_ CaCl2 + _2_ AgNO­3 _1_ Ca(NO3)2 + _2_ AgCl
How many grams of silver chloride are produced when 45 g of calcium chloride react with excess silver nitrate? = 45 g CaCl2 1 mol CaCl2 111 g CaCl2 0.405 mol CaCl2 0.811 mol AgCl 0.405 mol CaCl2 2 mol AgCl 1 mol CaCl2 g AgCl 0.811 mol AgCl 143.5 g AgCl 1 mol AgCl

__ C3H8 + _5_ O2­ _4_H2O + _3_CO2 How many moles of O2­ should be supplied to burn 1 mol of C3H8 (propane) molecules in a camping stove? (write down the equation first; this is a simple Combustion equation and remember that it must be balanced before you continue.) If I had asked for grams instead of molecules, then the last step would be: 1 mol C3H8 5 mol O2 = 5 mol O2 5 mol O2 6.022 × molecules O2 1 mol O2 = 𝟑.𝟎𝟏𝟏× 𝟏𝟎 𝟐𝟒 Molecules 5 mol O2 32 g O2 1 mol O2 = 160 g O2

_4_ Al + _3_ O _2_ Al2O3 Calculate the mass of alumina (Al2O3) produced when 100 g of aluminum burns in oxygen. (write down the equation first; this is a simple Decomposition equation and remember that it must be balanced before you continue.) = 3.7 mol Al = 1.85 mol Al2O3 = g Al2O3 100 g Al 1 mol Al 27 g Al 3.7 mol Al 2 mol Al2O3 4 mol Al 1.85 mol Al2O3 102 g Al2O3 1 mol Al2O3