Question

The reduction of 1.80g of a metal oxide required 833mL of ${{\text{H}}_{\text{2}}}$ 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

Hint:The gram equivalence of hydrogen will be equal to the gram equivalence of metal oxide. The reaction involved in the reduction of metal oxide using hydrogen is:
${{M}_{x}}{{O}_{y}}+y{{H}_{2}}\to xM+y{{H}_{2}}O$
The formulae of equivalent mass is as follows:
$Equivalent\,mass=\frac{Given\,mass}{Gram\,equivalent}$

Complete step-by-step answer:In the question it is asked that, in a reduction reaction, to reduce 1.8 g of metal oxide, 833ml of hydrogen was required and what will be the equivalent mass of metal oxide. Before going into the solution we should know what equivalent mass means.
Equivalent mass is defined as the mass of one equivalent i.e. that given quantity of mass which will either combine or displace the fixed quantity of another element. To get more understanding, we define an equivalent mass of an element as that weight of element which could displace 1.008 g H or 35.5g of Cl etc.
$Equivalent\,mass=\frac{Given\,mass}{Gram\,equivalent}$
And the gram equivalent is equal to the number of moles multiplied by n-factor.
$Gram\,equivalent=no.\,of\,moles\times n-factor$
The n- factor is defined as the total number of ${{H}^{+}}$ions in a compound taking part in the reaction.
Now let us solve the question. The reduction reaction of oxide with hydrogen taking place is as follows:
${{M}_{x}}{{O}_{y}}+y{{H}_{2}}\to xM+y{{H}_{2}}O$
And at STP condition, 1 mole of gas will be equal to 22.4L.
Now convert 883mL of hydrogen gas in terms of moles.
So 883mL of hydrogen is$=\frac{883}{22.4\times 1000}g=0.0372moles\,of\,{{H}_{2}}$
So now we know that 883mL of hydrogen gas is equal to 0.0372 moles of hydrogen.
Now we have to find an equivalent mass of hydrogen.
$Equivalent\,mass\,of\,{{H}_{2}}=No.\,of\,moles\,of\,{{H}_{2}}\times n$
Here the number of ${{H}^{+}}$ involved in the reaction is 2, so the n factor is 2.
$Equivalent\,mass\,of\,{{H}_{2}}=0.0372\times 2=0.0744g\,eq$
Since the equivalent weight of hydrogen will be equal to the equivalent weight of metal oxide.
$Equivalent\,weight=\frac{Given\,mass}{Equivalent\,mass}=\frac{1.8}{0.0744}=24.1935\approx 24.2g$

Hence the correct answer for the question is option (a) which 24.2g.

Note:We also refer to equivalent weight or equivalent mass as gram equivalence, both are the same and have the same meaning. Or we also refer to gram equivalent weight, students get often confused with the terms. Equivalent weight will not be the same for a species since it depends on the n-factor i.e. the number of${{H}^{+}}$ involved in the reaction.