Question

40 g of a sample of carbon on combustion left 10% of it unreacted. The volume of oxygen required at STP for this combustion reaction is

• A. 22.4 L
• B. 67.2 L
• C. 11.2 L
• D. 44.8 L

Answer 1

Answer: (B)

$\underset{12g}{C}+\underset{22.4L}{O_2}\to C{O}_2$ The amount of carbon unreacted $=40\times \frac{10}{100}g$ $=4g$ So, the amount of carbon reacted $=(40-4)g$ $=36g$ $\because$ At STP, for the combustion of 12 g of C, oxygen required is = 22.4 L $\therefore$ For the combustion of 36 g of C, oxygen required will be $=\frac{22.4}{12}\times 36L=67.2L$