Answer to Question #298846 in General Chemistry for Sydney

Question

When 50.1 g

g iron(III) oxide reacts with carbon monoxide, 31.7 g

g iron is produced. What is the percent yield of the reaction?Fe2O3(s)+3CO(g)→2Fe(s)+3CO2(g)

Answer 1

The molar masses are:

Fe₂O₃ - 159.7 g/mol

Fe - 55.85 g/mol

$Theoretical\ yield(Fe)=50.1\ g(Fe_2O_3)\times\frac{1\ mol(Fe_2O_3)}{159.7\ g(Fe_2O_3)}\times\frac{2\ mol(Fe)}{1 mol(Fe_2O_3)}\times\frac{55.85\ g(Fe)}{1\ mol(Fe)}=35.0\ g$

$\%\ yield=\frac{actual\ yield}{theoretical\ yield}\times100\%=\frac{31.7\ g}{35.0\ g}\times100\%=90.6\%$

Answer: 90.6%