Question

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

• A. 153.12 g of $KClO_3$
• B. 122.5 g of $KClO_3$
• C. 245 g of $KClO_3$
• D. 98.0 g of $KClO_3$

Answer 1

Answer: (A)

153.12 g of $KClO_3$

Answer 2

\underset{\text{2 \times122.5 g}}{ {2KClO_{3}}} \to 2KCl+\underset{\text{3\times 32 g}}{ {3O_{3}}} $3\times32\,g$ of $O_{2}$ 2 is produced from $KClO_{3}=245g$ $48\,g$ of $O_{2}$ is produced from $KClO_{3}=122.5\,g$ $\therefore 80\% KClO_{3}$ needed $=\frac{122.5 g\times 100}{80}$ $=153.12\,g$