Question

A metal oxide having formula $Z_2{O}_3$ gives free metal and water on reduction with hydrogen. $0.2$ g of the metal oxide can be completely reduced by $12$ mg of hydrogen. The atomic weight of the metal is:

• A. 12
• B. 13
• C. 26
• D. 78

Answer 1

Answer: (C)

26

$\underset{[2g=200mg]}{Z_2{O}_3}+\underset{\underset{=6mmol}{\frac{12mg}{2}}}{3H_2}\to 2Z+3H_{2}O$
Since $H_2$ used is $6$ mmol, $Z_2{O}_3$ used should be $2$ mmol.
Mol of $Z_2{O}_3=\frac{Weight}{Mw}=\frac{200mg\times {10}^{-3}g}{Mw}$ $=2\times {10}^{-3}moles$
$\therefore Mw\left(Z_2{O}_3\right)=100$
$\therefore 2Z+16\times 3=100$
$Z=\frac{100-48}{2}=26$ g