Question

A metal oxide has the formula $A_2{O}_3$. It can be reduced by hydrogen to give free metal and water. $0.1596g$ of this metal oxide requires $6mg$ of hydrogen for complete reduction. What is the atomic weight of metal?

• A. $52.3$
• B. $57.5$
• C. $55.8$
• D. $59.3$

Answer 1

The correct option is C $55.8$

$\underset{0.1596g}{A_2{O}_3}+\underset{0.006g}{3H_2}⟶2A+3H_{2}O$

$0.006g$ of $H_2$ reduces $0.1596g$ of $A_2{O}_3$

$6g$ of $H_2$ will reduces

$=\frac{0.01596\times 6}{0.006}=159.6g{A}_2{O}_3$

Hence, the molecular weight of $A_2{O}_3=159.6g$

Let the molecular weight of $A=x$

$\therefore$ $2x\times 3\times 16=159.6$

or $2x=159.6-48$

or $2x=111.6$

or $x=55.8$

So, the correct option is $C$