Question

A sample of ${KClO}_3$ on decomposition yielded $448mL$ of oxygen gas at STP. then the weight of ${KClO}_3$ originally taken was:

• A. $0.815g$
• B. $1.63g$
• C. $3.27g$
• D. $2.45g$

Answer 1

Answer: (B)

$1.63g$

$2KCl{O}_3⟶2KCl+3O_2$

Volume occupied by $1$ mole of oxygen $=22.4L$

$\therefore$ No. of moles of oxygen in $0.448L$ of oxygen $=\frac{1}{22.4}\times 0.448=0.02\text{moles}$

Molecular mass of $KCl{O}_3=122.5gm$

From the reaction,

Amount of $KCl{O}_3$ required to yield $3$ moles of oxygen $=245gm$

$\therefore$ Amount of $KCl{O}_3$ required to yield $0.02$ moles of oxygen $=\frac{245}{3}\times 0.02=1.63gm$

Hence, $1.63gm$ of $KCl{O}_3$ was originally taken.