Question

A metal oxide has the formula $M_2{O}_3$. It can be reduced by $H_2$ to give free metal and water. 0.15% g of $M_2{O}_3$ 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

The correct option is C 55.8

$\underset{(2x+48)g}{M_2{O}_3}+\underset{6g}{3H_2}\to 2M+3H_{2}O$

$x=$ Atomic mass of metal

$\therefore 0.006g{H}_2$ reduces 0.1596 g $M_2{O}_3$

$\therefore 6g{H}_2$ will reduce $\frac{0.1596}{0.006}\times 6g{M}_2{O}_3=159.6M_2{O}_3$

$2x+48=159.6$

$2x=111.6$

$x=55.8$

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