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How many liters of chlorine gas can react with 56.0 grams of calcium metal at standard temperature and pressure? Show all of the work used to find your answer. Ca + Cl2 CaCl2
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56 g of Ca -->1.40 mol Ca (divide grams by molar mass to get # of moles). Each mol of Ca reacts with one mol of Cl2, so you need 1.40 mol. @ stp, 1 mol = 22.4 L. So multiply 1.40 mol x 22.4 L/mol = 31.30 L of Cl2.
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