Question

Q. A metal oxide has the formula M2O3. If 0.1596g of the metal oxide requires 6 mg of hydrogen for complete reduction then atomic weight of the metal is--------?

Answer 1

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Answer 2

The atomic weight of metal in metal oxide is 55.8 gm

Explanation:

From the given question, we get the chemical equation as :

$M_2O_3 + H_2 \rightarrow 2M + 3H_2O$

We are also given that 6 mg of $H_2$ reduces 0.1596 gm of $M_2O_3$

Or

$6 \times 10^{-3} \ gm$ $H_2$ reduces 0.1596 gm of $M_2O_3$

So, 1 gm of $H_2$ will reduce $\frac{0.1596}{6 \times 10^{-3} gm}$ of $M_2O_3$  = 26.6 gm of $M_2O_3$

Equivalent weight of $M_2O_3$  = 26.6

Equivalent weight of M + Equivalent weight of O = 26.6

We know that equivalent weight of O = 8

So, Equivalent Weight of M + 8 = 26.6

Equivalent weight of M = 26.6 - 8 = 18.6

Valency of the metal in $M_2O_3$  is 3. This is because it forms an oxide of the form $M_2O_3$  and belongs to Group III in the periodic table. Group III elements have 13 electrons and 3 electrons in their valence shell making their valency as 3.  

$Equivalent \ Weight = \frac{Atomic \ Weight}{Valency}$

$18.6 = \frac{Atomic \ Weight}{3}$

So,

$Atomic \ Weight = 18.6 \times 3 = 55.8 \ gm$