SATHEE: Some Basic Concepts of Chemistry

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 $6 m g$ of hydrogen for complete reduction. The atomic weight of the metal is

(a) 27.9

(b) 159.6

(d) 55.8

[1989]

Topic 3: Stoichiometric Calculations

Show Answer

Solution:

(d) The reaction can be given as

$Z_{2} O_{3} + 3 H_{2} ⟶ 2 Z + 3 H_{2} O$

$0.1596 g$ of $Z_{2} O_{3}$ react with $H_{2} = 6 m g = 0.006 g$

$∴ 1 g$ of $H_{2}$ react with

$= \frac{0.1596}{0.006} = 26.6 g of Z_{2} O_{3}$

$∴$ 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 $= \frac{Atomic wt.}{valency}$

$∴$ At. wt. of $Z = 18.6 \times 3 = 55.8$

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Some Basic Concepts of Chemistry - Result Question 39

Topic 3: Stoichiometric Calculations

Solution:

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