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Aluminum reacts with oxygen to produce aluminum oxide. In the lab, you start with a 5.3 gram sample of aluminum and produce 3.48 grams of aluminum oxide. What is the percent yield? 4 Al + 3 O2 → 2 Al2O3 Al molar mass = 26.98 g/mol Al2O3 molar mass = 101.96 g/mol Report your answer with the correct number of significant figures.
Expert Answer
First, define % yield
% yiel = real amount obtained / theoretical amount expected * 100%
the theoretical amount can be claculated,
the real amount is measured , in this case is 3.48 g of Aluminium Oxide
now… clacualte theoretical yield
mol of Al= mass/MW = 5.3 / 26.98 = 0.1964 mol of Al
The reaction should:
4 mol of Al = 2 mol of Al2O3
0.1964 mol of Al = 2/4*0.1964 = 0.0982 mol of Al2O3
mass of Al2O3 = mol*MW = 0.0982*101.96 = 10.012472 g of Al2O3 are expected
therefore:
% yield = 3.48/10.012472 *100 = 34.75 %
correct sig fig:
2 sig fig in 5.3 g of Al, therefore, round p to 2 sig fig:
34.75 = 35 %
