Question

How much $KN{O}_3$ must be heated to give as much $O$, as required to burn 16 g of $S$ into $S{O}_2$?

$[N=14,O=16,K=39,S=32]$

[Hint : $2KN{O}_3,\to 2KN{O}_2+O_2;S+O_2\to S{O}_2$]

• A. $100g$
• B. $101g$
• C. $91g$
• D. $99.1g$

Answer 1

Answer: (B)

$101g$

$S+O_2⟶S{O}_2$

Moles of S $=\frac{16g}{32g/mol}=0.5mol$

0.5 mol of S requires 0.5 mol of oxygen.

So, we know that sulphur burns in 0.5 mol oxygen.

The thermal decomposition of Pottasium nitrate:

$2KN{O}_3⟶2KN{O}_2+O_2$

1 mol of $O_2$ is produced by 2 mol of $KN{O}_3$

$0.5mol$ of $O_2$ will be produced by 1 mol of $KN{O}_3$

Thus, 1 mole of $KN{O}_3$ is required to produce the given amount of oxygen.

1 mol of $KN{O}_3=$ molar mass of $KN{O}_3=101g$

Hence, option B is correct.