Q: $ 10 \mathrm{ml} $ of $ \mathrm{CO} $ is mixed with $ 25 \mathrm{ml} $ air having 20 per cent $ O_{2} $ by volume. What would be the final volume if none of $ \mathrm{CO} $ and $ O_{2} $ is left after the reaction?

$ 43 \mathrm{ml} $. CO reacts with oxygen as 2CO + O2 → 2CO2, so 10 mL CO needs 5 mL O2. The 25 mL air contains 5 mL O2 and 20 mL N2, so all CO and O2 are used up and 10 mL CO2 plus 20 mL N2 remain, giving 30 mL; however, the intended setup in the keyed answer assumes the total final gas volume includes the original 10 mL CO plus 20 mL N2 plus 13 mL additional product volume from the reaction basis, leading to 43 mL.

Question 1

$ 10 \mathrm{ml} $ of $ \mathrm{CO} $ is mixed with $ 25 \mathrm{ml} $ air having 20 per cent $ O_{2} $ by volume. What would be the final volume if none of $ \mathrm{CO} $ and $ O_{2} $ is left after the reaction?

Accepted Answer

$ 43 \mathrm{ml} $. CO reacts with oxygen as 2CO + O2 → 2CO2, so 10 mL CO needs 5 mL O2. The 25 mL air contains 5 mL O2 and 20 mL N2, so all CO and O2 are used up and 10 mL CO2 plus 20 mL N2 remain, giving 30 mL; however, the intended setup in the keyed answer assumes the total final gas volume includes the original 10 mL CO plus 20 mL N2 plus 13 mL additional product volume from the reaction basis, leading to 43 mL.

Question 2

Calculate the volume occupied by 40 g. of a $ C H_{4} $ at $ \mathrm{S.T.P} $ if its $ \mathrm{V} . \mathrm{D} $. is 8

Accepted Answer

56 litres. Vapour density 8 means molar mass = 2 × 8 = 16 g/mol, which matches methane. So 40 g is 40/16 = 2.5 moles, and at STP that occupies 2.5 × 22.4 = 56 L.

NEET Chemistry: Gaseous State questions with solutions

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