## Q. 1.1.1

Calculate the weight and volume of air required for the combustion of 3 kg of carbon.

## Verified Solution

Combustion reaction

$C \quad + \quad O_2 \rightarrow CO_2 \\12 kg \quad 32 kg$

Weight of oxygen required to burn 12 kg C = 32 kg

Weight of oxygen required to burn 3 kg     $C =\frac{32}{12} × 3 = 8 ~kg$

As, air contains 23% oxygen by weight

Therefore, weight of air required $= 8 × \frac{100}{23}= 34.783 ~kg = 34783 ~g$

Volume of air required

as, 1 mole of any gas at STP occupies 22.4 L

Therefore, volume occupied by 1 mole air = 28.94 g air = 22.4 L

(molecular weight of air = 28.94 g)

\begin{aligned}\text{Volume occupied by 34783 g of air}= & \frac{22.4}{28.94} × 34783\\ \\= & 26.92 × 10^3 ~L = 26.92 m^3\\ & (\text{Since} 1 ~L = 10^3 ~m^3)\end{aligned}