Question

when 45.0 g of Fe2O3 is reduced with excess H2 in a furnace, 2.60 g of metallic iron is recovered. what is the percent yield?

Answer 1

Molar mass of Fe2O3,

MM = 2*MM(Fe) + 3*MM(O)

= 2*55.85 + 3*16.0

= 159.7 g/mol

mass of Fe2O3 = 45 g

mol of Fe2O3 = (mass)/(molar mass)

= 45/1.597*10^2

= 0.2818 mol

Balanced chemical equation is:

Fe2O3 + 3H2 —> 2Fe + 3 H2O

According to balanced equation

mol of Fe formed = (2/1)* moles of Fe2O3

= (2/1)*0.2818

= 0.5636 mol

Molar mass of Fe = 55.85 g/mol

mass of Fe = number of mol * molar mass

= 0.5636*55.85

= 31.47 g

% yield = actual mass*100/theoretical mass

= 2.6*100/31.47

= 8.26 %

Answer: 8.26 %