Some Basic Concepts of Chemistry - Result Question 39
####39. 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 weight of the metal is
(a) 27.9
(b) 159.6
(c) 79.8
(d) 55.8
[1989]
Topic 3: Stoichiometric Calculations
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Solution:
- (d) The reaction can be given as
$Z_2 O_3+3 H_2 \longrightarrow 2 Z+3 H_2 O$
$0.1596 g$ of $Z_2 O_3$ react with $H_2=6 mg=0.006 g$
$\therefore 1 g$ of $H_2$ react with
$ =\frac{0.1596}{0.006}=26.6 g \text{ of } Z_2 O_3 $
$\therefore$ Eq. wt. of $Z_2 O_3=26.6$
(from the definition of eq. wt.)
Eq. wt. of $Z+$ Eq. wt. of $O(8)=26.6$
$\Rightarrow$ Eq. wt. of $Z=26.6-8=18.6$
Valency of metal in $Z_2 O_3=3$
Eq. wt.of metal $=\frac{\text{ Atomic wt. }}{\text{ valency }}$
$\therefore$ At. wt. of $Z=18.6 \times 3=55.8$
