Question

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.

Answer 1

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 %