Solution and explain

Solution and explain 130. The heat evolved during the combustion of 112 litre Of gas at STP (mixture Of volume of H2 arid CO) : Gitæn --241.8KJ COO + - coz(g) A H 283 FJ (1) 241.8 KJ (3) 1312 KJ (2) 283 KJ (4) 1586 KJ

Total heat evolved by reaction of H2(g) and CO(g) is: (241.8 + 283) kJ = 524.8 kJ (by Hess's law) Now volume occupied by a gas at STP is 22.4 L. Volume of H2(g) = volume of CO(g) = 112/2 L= 56 L Therefore , number of moles of each gas = 56/22.4 = 2.5 moles Heat evolved by reaction of one mole of each gas = 524.8 kJ Therefore , heat evolved by reaction of 2.5 moles of each gas = 524.8 *2.5 = 1312 kJ Hence , option (c) is correct. Heat evolved by reaction of one mole of each gas = 524.8 kJ