Question

If potassium chlorate is $80\%$ pure then $48g$ of oxygen would be produced from (atomic mass of $K=39u$)

• A. $153.12g$ of $KCl{O}_3$
• B. $122.5g$ of $KCl{O}_3$
• C. $245g$ of $KCl{O}_3$
• D. $98.0g$ of $KCl{O}_3$

Answer 1

The correct option is A $153.12g$ of $KCl{O}_3$

The corresponding reaction is,
$KCl{O}_3⟶KCl+\frac{3}{2}{O}_2$
So, from the balanced stoichiometric reaction, $1$ mole $KCl{O}_3$ produces $\frac{3}{2}$ moles of $O_2$.
Molar mass of $KCl{O}_3=122.5g$
Molar mass of $O_2=32g$
Hence, $122.5g$ $KCl{O}_3$ produces $48g$ of $O_2$.
But here percentage purity of potassium chlorate sample $KCl{O}_3$ is $80\%$
So, required amount of $KCl{O}_3=\frac{100\times 122.5}{80}g=153.12g$