Chemistry - Redox Reactions Question with Solution | TestHub

With NaOH, the valency factor of ${\mathrm{K}}_{\mathrm{x}}{\mathrm{H}}_{\mathrm{y}}({\mathrm{C}}_2{\mathrm{O}}_4{)}_{\mathrm{z}}\cdot {\mathrm{nH}}_2\mathrm{O}$ is y. Mili equivalents of salt = mili equivalents of NaOH.

$$\frac{2.54}{m}\times 10\times y=0.01\times 30$$

$M=$ molecular weight of salt. With ${\mathrm{KMnO}}_4$, the valency factor of ${\mathrm{K}}_{\mathrm{x}}{\mathrm{H}}_{\mathrm{y}}({\mathrm{C}}_2{\mathrm{O}}_4{)}_{\mathrm{z}}\cdot {\mathrm{nH}}_2\mathrm{O}$ is $2z$.

$$\frac{2.54}{m}\times 10\times 2z=0.01\times 40$$

$$\frac{\text{ eq. }1}{\text{ eq. }2}=\frac{y}{2z}=\frac{3}{4}$$

$$\frac{y}{z}=\frac{3}{2}$$

Apply charge balancing to ${\mathrm{K}}_{\mathrm{x}}{\mathrm{H}}_{\mathrm{y}}({\mathrm{C}}_2{\mathrm{O}}_4{)}_{\mathrm{z}}.{\mathrm{nH}}_2\mathrm{O}$:

$$x+y-2z=0(y=3,z=2)$$

$$x+3-4=0$$

$$x=1$$

${\mathrm{K}}_1{\mathrm{H}}_3({\mathrm{C}}_2{\mathrm{O}}_4{)}_2\cdot {\mathrm{nH}}_2\mathrm{O}$

Calculate 'm' from eq. 1:

$$m=254$$

$$39+3+88\times 2+18n=254$$

$$n=2$$

Hence, $x=1,y=3,z=2$