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 (Given: Atomic mass
of Fe = 56]
(1) 10.58%
(2) 20.28%
(3) 35.35%
(4) 11.11%​

Answer 1

Answer:11.11%

Explanation:

Answer 2

Answer:

The correct answer is option (4).

Explanation:

Mass of 1 mole of FeO = 72 g

$4FeO+O_2\rightarrow 2Fe_2O_3$

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

$\frac{2}{4} mol= 0.5 mol$ 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 :

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