Question

One mole sample of FeO is heated in air until it is completely converted into Fe2O3.
The percentage increase in weight of sample is ???

Answer 1

Answer:

The percentage  increase in weight of sample is 11.11%.

Explanation:

Mass of 1 mole of FeO = 72 g

$4FeO+O_2\rightarrow 2Fe_2O_3$

According to reaction 4 moles of FeO gives 2 moles of$Fe_2O_3$. Then 1 mole will give:

$\frac{2}{4} mol= 0.5 \text{mol of }Fe_2O_3$

Mass of 0.5 moles of Fe_2O_3: $0.5 mol\times 160 g/mol=80 g$

Change in mass = 80 g - 72 g = 8 g

The percentage  increase in weight of sample :

$\%=\frac{80 g-72 g}{72 g}\times 100=11.11\%$