Question

${{\text{H}}_2}{{\text{C}}_2}{{\text{O}}_4}.2\,{{\text{H}}_{\text{2}}}{\text{O}}$(mol. Wt = $126$) can be oxidised into ${\text{C}}{{\text{O}}_{\text{2}}}$by acidified ${\text{KMn}}{{\text{O}}_{\text{4}}}$. $6.3$gms of oxalic acid can be oxidised by
A. $3.16$ gms of ${\text{KMn}}{{\text{O}}_{\text{4}}}$
B. $200$ ml of $0.1$ M ${\text{KMn}}{{\text{O}}_{\text{4}}}$
C. $0.1$ mole of ${\text{KMn}}{{\text{O}}_{\text{4}}}$
D. $0.02$ moles of ${\text{KMn}}{{\text{O}}_{\text{4}}}$

Answer 1

Hint:To determine the number of moles and grams of any reactant or product balanced equation is required. We can determine the number of molecules of oxalic acid by using the mole formula.
After writing the balanced equation, by comparing the number of moles of oxalic acid and potassium permanganate from the balanced equation we can determine the moles of potassium permanganate required.

Complete solution:
First we will write the balanced chemical equation for the reaction of ${{\text{H}}_2}{{\text{C}}_2}{{\text{O}}_4}.2\,{{\text{H}}_{\text{2}}}{\text{O}}$ with ${\text{KMn}}{{\text{O}}_{\text{4}}}$.
$4\,{{\text{H}}_2}{{\text{C}}_2}{{\text{O}}_4}.2\,{{\text{H}}_{\text{2}}}{\text{O}}\, + \,2\,{\text{KMn}}{{\text{O}}_{\text{4}}}\, \to \,{{\text{K}}_2}{{\text{C}}_2}{{\text{O}}_4}.2\,{{\text{H}}_{\text{2}}}{\text{O}}\,{\text{ + }}\,2\,{\text{Mn}}{{\text{O}}_2}\, + \,10\,{{\text{H}}_{\text{2}}}{\text{O}}\,{\text{ + }}\,6\,{\text{C}}{{\text{O}}_{\text{2}}}$
$6.3$ gms of oxalic acid is getting oxidised and the molar mass of oxalic acid is$126$.
So, the mole of oxalic acid is,
${\text{Mole}}\,{\text{ = }}\,\dfrac{{{\text{Mass}}}}{{{\text{Molar}}\,{\text{mass}}}}$
On substituting $6.3$ g for mass and $126$for molar mass of oxalic acid,
${\text{Mole}}\,{\text{ = }}\,\dfrac{{6.3}}{{{\text{126}}}}$
${\text{Mole}}\,{\text{ = }}\,0.05$
According to the balanced equation, four moles of oxalic acid is getting oxidised by two moles of potassium permanganate so, $0.05$moles of oxalic acid will get oxidised by,
$4$ Mole of ${{\text{H}}_2}{{\text{C}}_2}{{\text{O}}_4}.2\,{{\text{H}}_{\text{2}}}{\text{O}}$= $2$mole of ${\text{KMn}}{{\text{O}}_{\text{4}}}$
$0.05$ Mole of ${{\text{H}}_2}{{\text{C}}_2}{{\text{O}}_4}.2\,{{\text{H}}_{\text{2}}}{\text{O}}$= $0.02$mole of ${\text{KMn}}{{\text{O}}_{\text{4}}}$
So, $6.3$gms of oxalic acid can be oxidised by $0.02$ moles of ${\text{KMn}}{{\text{O}}_{\text{4}}}$.
Therefore,the correct option is (B).

Note: Stoichiometry measurements are used to determine the amount of reactant or product from the given amounts. Stoichiometry measurements give quantitative relations among the amounts of various species of a reaction. To determine the stoichiometry relations a balanced equation is necessary. Percent yield is determined as the actual amount of product divided by the theoretical yield of product and multiplied with hundred. The molar mass of the compound is determined by adding the atomic mass of constituting atoms.