Question

Consider the following reaction:
xMn{O^ - }_2\; + {\text{ }}y{\text{ }}{C_2}{O_4}^{2 - } + {\text{ }}z{H^ + } \to \;\;xM{n^{2 + }} + {\text{ }}2yC{O_2} + \dfrac{z}{2}{H_2}{O_\;}
The values of x, y and z in the reaction are, respectively:
A- 2, 5 and 8
B- 2, 5 and 16
C- 5, 2 and 8
D- 5, 2 and 16

Answer 1

First of all balance the equation starting from the elements other than oxygen and hydrogen. Then count the x, y and z value.

Complete step by step solution:
Let’s start with assuming the minimum number of x i.e. 2, then equation becomes:
2Mn{O^ - }_2\; + {\text{ }}y{\text{ }}{C_2}{O_4}^{2 - } + {\text{ }}z{H^ + } \to \;\;2M{n^{2 + }} + {\text{ }}2yC{O_2} + \dfrac{z}{2}{H_2}{O_\;}
Now try to balance the oxygen of the equation for that we have two option 2 and 5 try with the higher one and find y:
2Mn{O^ - }_2\; + {\text{ 5 }}{C_2}{O_4}^{2 - } + {\text{ }}z{H^ + } \to \;\;2M{n^{2 + }} + {\text{ (}}2x5)C{O_2} + \dfrac{z}{2}{H_2}{O_\;}
2Mn{O^ - }_2\; + {\text{ 5 }}{C_2}{O_4}^{2 - } + {\text{ }}z{H^ + } \to \;\;2M{n^{2 + }} + {\text{ 10}}C{O_2} + \dfrac{z}{2}{H_2}{O_\;}
Count the oxygen atom in both side of equation and calculate the number of hydrogen to balance equation:
Oxygen in reactant side= Oxygen in product side
4+20 = 20+$\dfrac{z}{2}$
z = 8
2Mn{O^ - }_2\; + {\text{ 5 }}{C_2}{O_4}^{2 - } + {\text{ 8}}{H^ + } \to \;\;2M{n^{2 + }} + {\text{ 10}}C{O_2} + \dfrac{8}{2}{H_2}{O_\;}
2Mn{O^ - }_2\; + {\text{ 5}}{C_2}{O_4}^{2 - } + {\text{ 8}}{H^ + } \to \;\;2M{n^{2 + }} + {\text{ 10}}C{O_2} + 4{H_2}{O_\;}
After getting the equation check all elements in both side of equation:

Element	Reactant	Product
Manganese	2	2
Oxygen	24	24
Carbon	10	10
Hydrogen	8	8

So the reaction is balanced now. In this final balanced equation value of x, y and z is 2, 5 and 8 respectively.

Hence option (A) is correct.

Note: Always balance the hydrogen in the end. If no entry of hydrogen in reactant or in product side then add ${H^ + }$ in reactant side and ${H_2}O$ in the product side.