How many grams of aluminum oxide are produced according to the reaction below gi

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This is a limiting reactant problem where you have to find out which one is the limiting factor. While you can set up 2 separate stoichiometric calculations, and the one with the least/smallest result will be your answer. I, however, set up a mole ratio and find the limiting reactant FIRST, that way I only have to do one calculation/equation. 1. First, find the moles of each. Grams -> moles Given gram x 1 mol/molar mass Molar mass Al: 26.98 g/mol Molar mass O2: 32.00 g/mol 10.0 g Al x 1 mol Al/26.98 g Al = 0.3706 mol Al 19.0 g O2 x 1 mol O2/32.00 g O2 = 0.5938 mol O2 2. Then using the balanced equation, set up a mole ratio: Coefficients/Number of moles. 4 Al/0.3706 mol Al = 3 O2/x We pretend we don't know the moles of O2. By crossing multiplying, we find x = 0.2780 mol O2, which is less than 0.5938 mol O2. This tells us we will have EXTRA O2, and that Al is our limiting reactant. 3. Do a regular stoichiometry problem: Molar ratio: 2 mol Al2O3 : 4 mol Al .... which can be reduced to 1 mol Al2O3 : 2 mol Al Molar mass Al2O3: 101.96 g/mol 10.0 g Al x 1 mol Al/26.98 g Al x 1 mol Al2O3/2 mol Al x 101.96 g Al2O3/1 mol Al2O3 = 18.895 g Al2O3 Sig figs... 18.9 g Al2O3 Report Abuse

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