Question

If potassium chlorate is $80 \%$ pure then $48\, g$ of oxygen would be produced from:

• A. $153.12 \,g$ of $KClO _{3}$
• B. $120 \,g$ of $KClO _{3}$
• C. $20 \,g$ of $KClO _{3}$
• D. $90\, g$ of $KClO _{3}$

Answer 1

Answer: (A)

$ \underset{2 \times 122.5 \,g}{2 KClO _{3}} \rightarrow \underset{3 \times 32 \,g}{2 KCl +3 O _{2}}$

$3 \times 32 g$ of $O _{2}$ is produced from $KClO _{3}=245\,g$

$48 g$ of $O_2$, is produced from $KClO_{3}=245 \,g$

$ \therefore 80 \% KClO _{3}$ needed $=\frac{122.5 \times 100}{80}$

$=153.12\, g$