Question

The reduction of $1.80g$ of a metal oxide required $833mL$ of $H_2$ measured in standard conditions. Hence, equivalent mass of the metal oxide is:

• A. $24.2g$
• B. $12.1g$
• C. $48.4g$
• D. none of these

Answer 1

The correct option is D none of these

$M^{+2}$

$MO+H_2\to M+H_{2}O$

$M_{2}O+H_2\to 2M+H_{2}O$

1.8 g MO / 0.03719 mole = 48.4 g/mol of $MO$ or $M_{2}O$

M = 32.4 g/mol

$M^{+4}$

$M{O}_2+2H_2\to M+2H_{2}O$

1.8 g $M{O}_2/$ 0.01860 mole = 96.8 g/mol

M = 64.8 g/mol $

$M^{+3}$

$M_2{O}_3+3H_2\to 2M+3H_{2}O$

1.8 g $M_2{O}_3/0.01240=145$

M = 48.5 g/mol