Question

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

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

Answer 1

Answer: (A)

$\underset{2\times 122.5g}{2KCl{O}_3}\to 2KCl+\underset{3timesg}{3O_3}$
$3\times 32g$ of $O_{2}2$ is produced from $KCl{O}_3=245g$
$48g$ of $O_2$ is produced from $KCl{O}_3=122.5g$
$\therefore 80\%KCl{O}_3$ needed $=\frac{122.5g\times 100}{80}$
$=153.12g$