Question

What volume of oxygen at STP will be obtained by the action of heat on $20$g of $KCl{O}_3$? $[K=39,Cl=35.5,O=16]$

• A. 5.49 L
• B. 3.6 L
• C. 2.5 L
• D. 4.55 L

Answer 1

The correct option is A 5.49 L

The balanced equation is -

$2KCl{O}_3\to 2KCl+3O_2$

Molecular wt. of $KCl{O}_3=39+35.5+16\times 3=122.5$

No. of moles of $KCl{O}_3=\frac{20}{122.5}$

Now, $2$ moles of $KCl{O}_3$ gives $3\times 22.4=67.2L$ of $O_2$

$⟹$ $\frac{20}{122.5}$ moles of $KCl{O}_3$ gives $\frac{20}{122.5}\times \frac{67.2}{2}L$ of $O_2$

$=5.49L$

Answer-(A)