Question

1.225 g sample of $KCl{O}_3$ was heated under such conditions that a part of it decomposed according to the equations given below. If the amount of $O_2$ evolved was 168 mL at STP, calculate the weight of $KCl{O}_4$.
$\text{(Molar mass of}KCl{O}_3=122.5gmo{l}^{-1},KCl{O}_4=138.5gmo{l}^{-1})KCl{O}_3\to KCl+O_{2}KCl{O}_3\to KCl{O}_4+KCl$

• A. 0.875 g
• B. 0.519 g
• C. 0.250 g
• D. 0.315 g

Answer 1

The correct option is B 0.519 g

$KCl{O}_3\to KCl+O_2$
Applying the principle of atom conservation (POAC) to O,
Moles of oxygen in $KCl{O}_3$ = Moles of oxygen in $O_2$
$3\times \text{moles of}KCl{O}_3=2\text{moles of}{O}_2$
$3\times \frac{WeightofKCl{O}_3}{molarmass}=2\times \frac{Volumeof{O}_2\left(mL\right)atSTP}{22400}$
$3\times \frac{WeightofKCl{O}_3}{122.5}=2\times \frac{168}{22400}$
Weight of $KCl{O}_3$ = 0.6125 g
The amount of $KCl{O}_3$ left = 1.225 – 0.6125 = 0.6125 g
$KCl{O}_3\to KCl{O}_4+KCl$
Applying the principle of atom conservation (POAC) to O,
Moles of oxygen in $KCl{O}_3$ = Moles of oxygen in $KCl{O}_4$
3 $\times$ moles of $KCl{O}_3$ = 4 × moles of $KCl{O}_4$
$3\times \frac{WeightofKCl{O}_3}{molarmass}=4\times \frac{WeightofKCl{O}_4}{molarmass}3\times \frac{0.6125}{122.5}=4\times \frac{WeightofKCl{O}_4}{138.5}WeightofKCl{O}_4=0.519g$