If you start with 150g of Fe2O3 as the limiting reactant, what is the theoretical yield of Fe?

1) (a) If you start with 150 g of Fe2O3 as the limiting reactant, what is the theoretical yield of Fe? Ans = 104.95g

(b) If your actual yield is 87.9 g, what is the percent yield? Ans = 83.75%

Fe2O3(s)+ 3 CO(g) > 2 Fe (s) + 3 CO2(g)

Explanation:

Fe = 55.8 g/mol

O = 16.0 g/mol

Molar mass of Fe2CO3 = (55.8*2) + (16.0*3)

Molar mass of Fe2CO3 = 159.6 g/mol

150g of Fe2O3 contains x moles

159.6g contains 1 mole

150*1 / 159.6 = 0.9398 moles

Number of participating moles in reaction = 2

Therefore; 2 * 0.9398 = 1.8796 moles are in the reaction.

Number of moles = mass / molar mass

Mass = number of moles * molar mass

Mass = 1.8796 * 55.84 = 104.95g of Fe.

104.95g of Fe is theoretical yield.

B. % yield = (actual yield / theoretical yield) *100

Actual yield = 87.9g

% yield = (87.9 / 104.95) * 100

% yield = 83.75%

the awnser is group 12