Question

A metal oxide has the formula $X_2{O}_3$. It can be reduced by hydrogen to give free metal and water. $0.1596g$ of metal oxide requires $6mg$ of hydrogen for complete reduction. The atomic mass of metal in amu is

• A. $15.58$
• B. $155.8$
• C. $5.58$
• D. $55.8$

Answer 1

Answer: (D)

$55.8$

We have,

$X_2{O}_3+3H_2\to 2X+3H_{2}O$

so, moles of $X_2{O}_3$ = $3\times$moles of $H_2$

moles of $H_2$ = $\frac{6}{2}=3$

so, $\frac{0.1596}{M}=9$

where M is a molecular mass of $X_2{O}_3$

Molecular mass of $X_2{O}_3$ = 2$\times$mass of metal + 48

so, on solving both equation we have

mass of metal $=55.8amu$