Question

Consider this reaction, which occurs in the atmosphere and contributes to photochemical smog: 4Fe(s) + 3O2(g) →2Fe2O3(s) If there is 15.87 g Fe and excess O2 present, the reaction yields 18.2 g Fe2O3. Calculate the percent yield for the reaction.

Answer 1

Molar mass of Fe = 55.85 g/mol

mass of Fe = 15.87 g

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

= 15.87/55.85

= 0.2842 mol

According to balanced equation

mol of Fe2O3 formed = (2/4)* moles of Fe

= (2/4)*0.2842

= 0.1421 mol

Molar mass of Fe2O3,

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

= 2*55.85 + 3*16.0

= 159.7 g/mol

mass of Fe2O3 = number of mol * molar mass

= 0.1421*1.597*10^2

= 22.69 g

% yield = actual mass*100/theoretical mass

= 18.2*100/22.69

= 80.2 %

Answer: 80.2 %