Percent yield

Question

How many grams of aluminum  would be produced if this reaction had a percent yield of 95% and you start with 9.73 grams of aluminum oxide reacton with an excess of carbon?

Answer 1

2Al2O3 + C ===> 4Al(s) + 3CO2  balanced equation

moles of Al2O3 = 9,73 g x 1 mol/102 g = 0.0954 moles

Mole ratio of Al : Al2O3 is 4 : 2 or 2:1

Theoretical yield of Al = 4 mol Al/2 moles Al2O3 x 0.0954 moles AlO3 = 0.1908 moles of Al theortical

Actual yield = 0.95 x 0.1908 moles = 0.181 moles Al

mass of Al = 0.181 moles Al x 27 g/mol = 4.89 grams (to 3 significant figures)