Question

A metal oxide has the formula $Z_2{O}_3$. It can be reduced by hydrogen to give free metal and water. 0.1596 g of the metal oxide requires 6 mg of hydrogen for complete reduction. The atomic mass of the metal is:

• A. 27.9
• B. 159.6
• C. 79.8
• D. 55.8

Answer 1

Answer: (D)

55.8

$Z_2{O}_3+3H_2\to 2Z+3H_{2}O$

1 mole of the metal requires 3 moles of $H_2$ for complete reduction.

Moles of $H_2$ consumed $=\frac{6\text{mg}}{2\text{g}}=3\times {10}^{-3}\text{mole}$

Moles of metal oxide reduced = $\frac{3\times {10}^{-3}\text{mole}}{3}={10}^{-3}\text{mole}$

$\therefore$ molecular mass of $Z_2{O}_3=\frac{0.1596\text{g}}{{10}^{-3}}=159.6\text{g}$

$\Rightarrow 2\times M_Z+3\times M_O=159.6$

$\Rightarrow 2\times M_Z+3\times 16=159.6$

$\Rightarrow M_Z=\frac{(159.6-48)}{2}=55.8$