Question

A metal oxide has the formula $A_{2}O_{3}$ . It can be reduced by hydrogen to give free metal and water. $0.1596\, g$ of this metal oxide requires $6\, mg$ 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

Answer: (C)

55.8

Answer 2

$\underset{0.1596\, g}{A _{2} O _{3}}+\underset{0.006\, g}{3 H _{2}} \longrightarrow 2 A +3 H _{2} O$ $0.006\, g$ of $H _{2}$ reduces $0.1596\, g$ of $A _{2} O _{3}$ $6\, g$ of $H _{2}$ will reduces $=\frac{0.01596 \times 6}{0.006}=159.6\, g A _{2} O _{3}$ Hence, molecular weight of $A_{2} O _{3}=159.6\, g$ Let molecular weight of $A=x$ $\therefore 2 x\times 3 \times 16=159.6$ or $2 x =159.6-48$ or $2 x =111.6$ or $x =55.8$