Step 1
The reaction must be written and balanced:
4Cs + O2 → 2Cs2O
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Step 2
Determine the limiting reactant:
The molar mass Cs) 133 g/mol
The molar mass O2) 32 g/mol
4Cs + O2 → 2Cs2O
Procedure:
4 x 133 g Cs ------ 32 g O2
46.1 g Cs ------ X = 2.77 g O2 is needed
("----" means 4 x 133 grams of Cs reacts to 32 g of O2)
For 46.1 g Cs, 2.77 g O2 is needed, but there is 13.4 g O2. Therefore, the limiting reactant is Cs.
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Step 3
Determine the theoretical yield of Cs2O from Cs:
4Cs + O2 → 2Cs2O
The molar mass Cs2O) 282 g/mol
Procedure:
4 x 133 g Cs ----- 2 x 282 g/mol Cs2O
46.1 g Cs ----- X = 48.8 g Cs2O = theoretical yield
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Step 4
% yield is defined as = actual yield/theoretical yield x 100
The actual yield is already provided = 28.3 g Cs2O
So, % yield = 28.3 g/48.8 g x 100 = 58 % approx.
Answer: % yield = 58 %