60 ml of a mixture of nitrous oxide N 2 O and nitric oxide ( NO ) was exploded with excess of hydrogen. If 38 ml of N 2 was formed, calculate the volume of each gas in the mixture.

Both nitrous oxide and nitric oxide react with hydrogen to produce hydrogen gas and water vapour. The reactions can be represented as, 2 NO + 2 H 2 → N 2 + 2 H 2 O N 2 O + H 2 → N 2 + H 2 O Suppose the volume of NO in the mixture is ' 2 x ' ml and N 2 O is ' 60 - 2 x ' ml . 2 NO + 2 H 2 → N 2 + 2 x x 2 H 2 O N 2 O + H 2 → N 2 + H 2 O 60 - 2 x 60 - 2 x Total volume of N 2 formed = 60 - 2 x + x According to the question, 60 - 2 x + x = 38 x = 22 ml Hence, volume of NO in the mixture = 2 × 22 = 44 ml , and volume of N 2 O in the mixture = 60 - 44 = 16 ml .

A sample of gaseous hydrocarbon occupying 1 . 12 litre at NTP , completely burnt in air and produced 2 . 2 g of CO 2 and 1 . 8 g H 2 O . Calculate the mass of compound taken and volume of O 2 at NTP required for its burning. Also calculate molecular formula of hydrocarbon.

1 . 12 litre of hydrocarbon occupies mole = 1 . 12 22 . 4 = 0 . 05 Mole of CO 2 formed = 2 . 2 44 = 0 . 05 Mole of H 2 O formed = 1 . 8 18 = 0 . 10 Let the hydrocarbon be C a H b C a H b + a + b 4 O 2 ⟶ aCO 2 + b 2 H 2 O ∴ 1 mole of C a H b gives = a mole CO 2 ∴ 0 . 05 mole of C a H b will give = 0 . 05 a mole of CO 2 ∴ 0 . 05 a = 0 . 05 ∴ a = 1 ∴ 1 mole of C a H b gives = b 2 mole H 2 O ∴ 0 . 05 mole of C a H b will give = 0 . 05 × b 2 mole H 2 O = 0 . 05 × b 2 = 0 . 10 ∴ b = 4 ∴ Hydrocarbon is CH 4 . ∴ Mass of 1 . 12 litre CH 4 at STP = 16 × 1 . 12 22 . 4 = 0 . 8 g Also, CH 4 + 2 O 2 ⟶ CO 2 + 2 H 2 O mole CH 4 requires = 2 × 32 gO 2 ∴ 0 . 05 mole CH 4 requires = 2 × 32 × 0 . 05 = 3 . 2 gO 2 = 2 . 24 litre O 2