Question

Potassium chlorate is 75% pure. 48gm. of oxygen would be produced from approximately how much potassium chlorate? (K = 39).

• A. 153.12 gm.
• B. 163.33 gm.
• C. 122.5 gm.
• D. 98.0 gm.

Answer 1

The correct option is B 163.33 gm.

$KCl{O}_3\stackrel{\Delta }{\to }KCl+\frac{3}{2}{O}_2$

$48gm{O}_2=\frac{48}{32}mol{O}_2$

$=1.5mol{O}_2$

$1.5mol{O}_2$ require $1molKCl{O}_3$

i.e, $(40+35.5+48)gKCl{O}_{3}123.5gKCl{O}_3$

$75\%$ is pure

$\therefore 75\%$ of $x=123.5$

$\Rightarrow x=\frac{123.5\times 100}{48}=\frac{494}{3}$

$=164.67g$