Question

An iron nail rusts when exposed to oxygen. For the following reaction, 5.55 grams of iron are mixed with excess oxygen gas . The reaction yields 7.32 grams of iron(III) oxide . iron ( s ) + oxygen ( g ) iron(III) oxide ( s ) What is the theoretical yield of iron(III) oxide ? grams What is the percent yield for this reaction ? %

Answer 1

Balanced reaction is

2Fe ( s ) + 3 O₂ ( g ). $\rightarrow$   Fe₂O₃ ( s )

Moles of Iron ( Fe ) = weight in gm / molar mass = 5.55 gm / 55.845 gm mol^—1 = 0.099 mol

moles of Fe₂O_{3 =} 1/2 x mol of Fe = 0.099/2 = 0.05 mol

weight of Fe₂O₃ = mol x molar mass = 0.05 mol x 159.7 gm mol^-1 = 7.985 gm

so theoretical yield = 7.985 gm

% yield = ( 7.32 gm / 7.985 gm ) x 100 = 91.67 %