Mass Limiting Reactant Problem 

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Presentation transcript:

Mass Limiting Reactant Problem  How much molten iron is formed from the reaction of 25.0 g FeO and 25.0 g Al? 3 FeO(l) + 2 Al(l) → 3 Fe(l) + Al2O3(s) First, let’s convert g FeO to g Fe: We can produce 19.4 g Fe if FeO is limiting. 25.0 g FeO × 3 mol Fe 3 mol FeO 1 mol FeO 71.85 g FeO × 1 mol Fe 55.85 g Fe = 19.4 g Fe Chapter 10

Problem, continued Second, lets convert g Al to g Fe: 3 FeO(l) + 2 Al(l) → 3 Fe(l) + Al2O3(s) Second, lets convert g Al to g Fe: We can produce 77.6 g Fe if Al is limiting. 25.0 g Al × 3 mol Fe 2 mol Al 1 mol Al 26.98 g Al × 1 mol Fe 55.85 g Fe = 77.6 g Fe Chapter 10

Problem Finished Let’s compare the two reactants: 25.0 g FeO can produce 19.4 g Fe 25.0 g Al can produce 77.6 g Fe FeO is the limiting reactant. Al is the excess reactant. Chapter 10