Question

The volume of CO₂ liberated at STP on burning 24 g of carbon in excess oxygen is_____

• A. 22.4 L
• B. 44.8 L
• C. 16.8 L
• D. 67.2 L

Answer 1

The correct option is B

44.8 L

$$\begin{gathered} \mathrm{C}\left(\mathrm{s}\right)+{\mathrm{O}}_2\left(\mathrm{g}\right)\stackrel{∆}{\to }{\mathrm{CO}}_2\left(\mathrm{g}\right) \\ \left(\mathrm{Carbon}\right)\left(\mathrm{Oxygen}\right)(\mathrm{Carbon}\mathrm{dioxide}) \end{gathered}$$

The molecular weight of carbon dioxide (CO₂)= 44 g

According to the mole concept, 1 mole of a substance= 22.4 L of the gas at STP. Therefore, from the balanced chemical equation, it is evident that,

12 g of carbon reacts with oxygen to form= 22.4 L of carbon dioxide

1 g carbon reacts with oxygen to form= $\frac{22.4}{12}$ L of carbon dioxide

24 g carbon reacts with oxygen to form= $\frac{22.4}{12}\times 24=44.8$ L of carbon dioxide

Hence option (b) is correct.