A 100. mL sample of 0.200 M aqueous hydrochloric acid is added to 100. mL of 0.200 M aqueous ammonia in a calorimeter whose heat capacity (excluding any water) is 480. J/K. The following reaction occurs when the two solutions are mixed. HCl(aq) + NH3(aq) → NH4Cl(aq) The temperature increase is 2.34°C. Calculate ΔH per mole of HCl and NH3 reacted.

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Question:

A 100. mL sample of 0.200 M aqueous hydrochloric acid is added to 100. mL of 0.200 M aqueous ammonia in a calorimeter whose heat capacity (excluding any water) is 480. J/K. The following reaction occurs when the two solutions are mixed. HCl(aq) + NH3(aq) → NH4Cl(aq) The temperature increase is 2.34°C. Calculate ΔH per mole of HCl and NH3 reacted.

Answers

Answer:-154KJ/molExplanation:mole of 100ml sample of 0.2M aqueous HCl = Molarity × volume in Liter = 0.2 × 100 / 1000  ( 1L = 1000 ml) = 0.02 mol and 0.02 mole of HCl solution require 0.02 mole of ammonia according to the mole ratio in the balanced equation.Heat loss by the reaction = heat gain by calorimeter = mcΔT + 480 J/Kwhere m is the mass of water = 100g + 100g = 200g since mass of 100ml of water = 100g and it is in both of them and specific heat capacity of water 4.184 J/gKheat gain by calorimeter  = (4.184 × 200 + 480) × 2.34 = 3081.3 JΔH per mole = heat loss / number of mole = 3081.3 / 0.02 = 154065.6 = -154KJ/mol

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