Question

A metal oxide has the formula $M_2O_3$. It can be reduced by $H_2$ to give free metal and water. $0.1596\, g$ of $M_2O_3$ required 6 mg of $H_2$ for complete reduction. The atomic mass of the metal is

• A. $27.9$
• B. $79.8$
• C. $55.8$
• D. $159.8$

Answer 1

Answer: (C)

$55.8$

Answer 2

$\underset{\text{(2x + 48) g}}{{M2O3}}+$ $\underset{\text{6 g}}{{3H2}}$ ${->2M +3H2O}$
$x =$ atomic mass of metal
$\because 0.006 \,g H_{2}$ reduces $0.1596 \,g M_{2}O_{3}$
$\therefore 6\,g \,H_{2}$ will reduce $\frac{0.1596}{0.006}\times6 \,g6 \,M_{2}O_{3}=159.66 \,M_{2}O_{3}$
$2x + 48 = 159.6 \Rightarrow 2x=111.6 \Rightarrow x=55.8$