Question

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Suppose 160 mL of 8.0 M NaOH(aq) is added to 197 mL of 6.0 M HNO3(aq) to make a solution with a density of 1.1 g/mL and a specific heat of 4.1 J/gxK. If the initial temperature of both solutions is 295 K, what will be the final temperature (K)? Enter your answer as an integer. H+(aq) + OH-(aq) → H2O() ΔHo=-55.8 kJ/mol

Answer 1

Initial moles of H+ 197 mL× 1000 mL 1L lmol HNO, 1.182 molH 8.0 mol NaOH 1L 1mol OH 1mol NaOH Initial moles of OH- 160 mL×ーー!× 1000 mL = 1.280 mol OH The two reactants are in equal amounts that are in stoichiometric proportions as per reaction. So the moles of water produced from either of initial moles will be x 1mol H2O = 1.182mol H2O lmol H 1.280 mol OH-x Imol F 1.280 mol H,o lmolOH Since HNO, acts as the limiting reactant, 1.182 moles of H2O produced 55.8 kJ Enthalpy of neutralization1182 mol H2Ox mol H2O 65.9556 kJ heat absorbed , qçalar -(-65.9556 k) 65.9556 kJ 65955.6 J Volume of water = (160 + 197) mL =357 mL The heat absorbed by calorimeter contents (water )is 1610.07· 659556 J = -295 K) T2-295 K = 6595561 1610.07J 7- 41K + 295 K- 336K