Question

$Mn{O}_2+2KCl{O}_3\to K_{2}Mn{O}_4+C{l}_2+2O_2$

$C{l}_2+K_{2}Mn{O}_4\to 2KCl+Mn{O}_2+O_2$

Each reaction takes place to the extent of 50 %.

11.2 L of $O_2$ at STP is obtained from $KCl{O}_3$ using?

• A. 0.66 mol
• B. 1.12 mol
• C. 1.33 mol
• D. 2.66 mol

Answer 1

The correct option is A 0.66 mol

$Mn{O}_2+2KCl{O}_3⟶K_{2}Mn{O}_4+C{l}_2+2O_2⟶\left(1\right)$

$C{l}_2+K_{2}Mn{O}_4⟶2KCl+Mn{O}_2+O_2⟶\left(2\right)$

At STP $11.2L$ of $O_2=\frac{11.2}{22.4}$ mole of $O_2=0.5$ mole

Reactions $1$ and $2$ give two mole and one mole of $O_2$ respectively.

Reaction $1$ gives $\frac{2}{3}\times 0.5=0.33$ mole of $O_2$.

Again each reaction takes place to the extent of $50$%.

$\therefore$ Number of moles of $KCl{O}_3=\frac{0.33}{0.5}=0.66$ mol