Chemistry - Redox Reactions Question with Solution | TestHub

First, balance the given equation:

$${\mathrm{K}}_2{\mathrm{Cr}}_2{\mathrm{O}}_7+14\mathrm{HCl}\to 2\mathrm{KCl}+2{\mathrm{CrCl}}_3+3{\mathrm{Cl}}_2+7{\mathrm{H}}_2\mathrm{O}$$

From the balanced equation, 1 mole of ${\mathrm{K}}_2{\mathrm{Cr}}_2{\mathrm{O}}_7$ produces 3 moles of ${\mathrm{Cl}}_2$.

Now, consider the reaction of ${\mathrm{MnO}}_2$ with $\mathrm{HCl}$:

$${\mathrm{MnO}}_2+4\mathrm{HCl}\to {\mathrm{MnCl}}_2+{\mathrm{Cl}}_2+2{\mathrm{H}}_2\mathrm{O}$$

From this balanced equation, 1 mole of ${\mathrm{MnO}}_2$ produces 1 mole of ${\mathrm{Cl}}_2$.

Since 1 mole of ${\mathrm{K}}_2{\mathrm{Cr}}_2{\mathrm{O}}_7$ produces 3 moles of ${\mathrm{Cl}}_2$, we need 3 moles of ${\mathrm{MnO}}_2$ to produce the same amount of ${\mathrm{Cl}}_2$.

The molar mass of ${\mathrm{MnO}}_2$ is:

$55+2\times 16=55+32=87g/mol$

Therefore, the mass of 3 moles of ${\mathrm{MnO}}_2$ is:

$3\times 87=261\mathrm{g}$

Now, let's find out how many moles of ${\mathrm{K}}_2{\mathrm{Cr}}_2{\mathrm{O}}_7$ react with 1 mole of HCl. From the balanced equation:

$${\mathrm{K}}_2{\mathrm{Cr}}_2{\mathrm{O}}_7+14\mathrm{HCl}\to 2\mathrm{KCl}+2{\mathrm{CrCl}}_3+3{\mathrm{Cl}}_2+7{\mathrm{H}}_2\mathrm{O}$$

14 moles of HCl react with 1 mole of ${\mathrm{K}}_2{\mathrm{Cr}}_2{\mathrm{O}}_7$. Therefore, 1 mole of HCl reacts with $\frac{1}{14}$ moles of ${\mathrm{K}}_2{\mathrm{Cr}}_2{\mathrm{O}}_7$.

The amount of ${\mathrm{Cl}}_2$ produced by $\frac{1}{14}$ moles of ${\mathrm{K}}_2{\mathrm{Cr}}_2{\mathrm{O}}_7$ is $3\times \frac{1}{14}=\frac{3}{14}$ moles.

To produce the same amount of ${\mathrm{Cl}}_2$ ($\frac{3}{14}$ moles) from ${\mathrm{MnO}}_2$, we need $\frac{3}{14}$ moles of ${\mathrm{MnO}}_2$.

Mass of ${\mathrm{MnO}}_2$ required = $\frac{3}{14}\times 87=\frac{261}{14}=18.642\mathrm{g}$