Question

Volume of co2 liberated at STP on burning 24g of methane in excess of air is 1) 11.2L 2) 22.4L 3) 33.6L 4) 44.8L

Answer 1

The correct option is (3). The explanation is given below.

Answer 2

Answer:

The correct answer is 3) 33.6 L.

Explanation:

The formula for CO2 liberated at STP on the burning of methane in excess of air is as follows:

$\mathrm{CH}_{4}+2 \mathrm{O}_{2}=\mathrm{CO}_{2}+2 \mathrm{H}_{2} \mathrm{O}$

The Molar mass of methane = Mass of C + 4 x (Mass of H)

= [12 + 4 x (1)] g mol

= (12 + 4) g mol

16 g of $\mathrm{CH}_{4}$ gives 22.4 L of $\mathrm{CO}_{2}$ at STP

24 g of $\mathrm{CH}_{4}$ gives $22.4 \times \frac{24}{16}=33.6 \mathrm{L}$ of $\mathrm{CO}_{2}$ at STP.