Question

If 42.6 grams of Al reacts completely with O2 and 57.8 grams of Al2O3 produced, what is the % yield of Al2O3for the reaction? 4Al + 3O2 --------> 2Al2O3Group of answer choices22.3%65.9%55.7%75.2%71.8%

Answer 1

Step 1

Data provided:

The reaction: 4Al + 3O2 => 2Al2O3 (balanced)

The limiting reactant Al (it is said that Al reacts completely)

Mass of Al = 42.6 g

The actual yield = 57.8 g Al2O3

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Step 2

Data needed: the molar mass of Al (26.9 g/mol) and Al2O3 (101.9 g/mol)

The theoretical yield:

4 x 26.9 g Al ------- 2 x 101.9 g Al2O3

42.6 g Al ------- X = 80.7 g Al2O3 = theoretical yield

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Step 3

% yield = (actual yield/theoretical yield) x 100 = (57.8 g/80.7 g) x 100 = 71.6 %

(this is the closest value for one of the options, 71.8 %)

Answer: 71.8 %