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1 December, 08:21

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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  1. 1 December, 08:44
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    -154KJ/mol

    Explanation:

    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/K

    where 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/gK

    heat gain by calorimeter = (4.184 * 200 + 480) * 2.34 = 3081.3 J

    ΔH per mole = heat loss / number of mole = 3081.3 / 0.02 = 154065.6 = - 154KJ/mol
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