Question

44.8 L of $O_2$ gas is collected at STP. If 4 mol of total $KCl{O}_3$ is used for the following reactions:
$KCl{O}_3\left(s\right)⟶KCl\left(s\right)+O_2\left(g\right)$
$KCl{O}_3⟶KCl{O}_4+KCl$
the number of moles of $KCl{O}_4$ formed woudl be:

• A. $2mol$
• B. $\frac{4}{3}mol$
• C. $\frac{1}{4}mol$
• D. $6mol$

Answer 1

The correct option is A $2mol$

Moles of Oxygen gas produced
$=\frac{\text{Volume(L) at STP}}{22.4L}=2mol$
In a chemical reaction, atoms are conserved, so moles of atoms shall also be conserved. This is known as principal of atomic conservation (POAC). This principal is helpful in solving problems of nearly all stoichiometric calculations.
Applying POAC for "O" atom
$Molesof"O"atomin{O}_2=Molesof"O"atominKCl{O}_3$
$2\times n_{O_2}=3\times n_{KCl{O}_3}$
$n_{KCl{O}_3}=\frac{4}{3}mol$
So $4-\frac{4}{3}$ moles remain for the second reaction.
For second reaction
$3\times n_{KCl{O}_3}=4\times n_{KCl{O}_4}$
$3\times \frac{8}{3}=4\times n_{KCl{O}_4}$
$n_{KCl{O}_4}=2mol$