Question

One mole sample of $FeO$ is heated in air until it is completely converted into $F{e}_2{O}_3$. The percentage increase in weight of the sample is?

[Given: Atomic mass of Fe = 56]

• A. $10.58\%$
• B. $20.28\%$
• C. $35.35\%$
• D. $11.11\%$

Answer 1

Answer: (D)

$11.11\%$

Mass of $1$ mol of $FeO=(56+16)=72g$

$FeO+\frac{3}{4}{O}_2⟶\frac{1}{2}F{e}_2{O}_3$

$1$ mol of $FeO$ gives $\frac{1}{2}$ mol $F{e}_2{O}_3$.

$\therefore$ Mass of sample after conversion $=\frac{1}{2}\times M_{F{e}_2{O}_3}=\frac{1}{2}\times 160=80g$

$\therefore \%$ increase in weight $=\frac{80-72}{72}\times 100=11.11\%$

Hence, the correct option is $\text{D}$