Question

The volume of oxygen at STP required to burn 2.4 g of carbon completely is

• A. 1.12 L
• B. 8.96L
• C. 2.24 L
• D. 4.48L

Answer 1

The correct option is D

4.48L

C(s) +$O_2$ (g)$\underset{}{\overset{}{\to }}$(g) CO${}_2$(g)

moles = 1mole 1mole 1mole

weight = 12gm 32gm 44gm

12gm of C require $\to$1mole of $O_2$

$\therefore$ 2.4gm of C will require $\to \frac{1}{12}\times 2.4$ mole of $O_2$

volume of $\frac{2.4}{12}$ mole $O_2$ at STP $=\frac{22.4\times 2.4}{12}$litre

$4.48$ litre