You react 100.0 ml of a 1.01 M HBr(aq) solution with 99.1 mlof a 1.05 M KOH(aq)

Question

You react 100.0 ml of a 1.01 M HBr(aq) solution with 99.1 mlof a 1.05 M KOH(aq) solution. The density of each solution is 1.04g/ml. The heat capacity of the calorimeter is 7.0 J/C. You observea T of 6.3 C. Assume a 4.18 J/g C specific heat for thesolution. Calculate the heat evolved in the neutralization. Use thecorrect Sign. Calculate the molar heat of neutralization based on the heatevolved. You react 100.0 ml of a 1.01 M HBr(aq) solution with 99.1 mlof a 1.05 M KOH(aq) solution. The density of each solution is 1.04g/ml. The heat capacity of the calorimeter is 7.0 J/C. You observea T of 6.3 C. Assume a 4.18 J/g C specific heat for thesolution. Calculate the heat evolved in the neutralization. Use thecorrect Sign. Calculate the molar heat of neutralization based on the heatevolved.

Explanation / Answer

We Know that :       KOH + HBr --------> KBr + H2O     0.101         0.103               0.101          We Know that :             Q = m x s x t + 7.0 J / C x 6.3 0C                    = 0.101 moles x 119 gm / mole x  4.18 J/ gm C x 6.3 0 C  +   7.0 J / C x 6.30 C                     = - 360.608 J           The heat is evolved during the process of Neutralizationreaction.           Molarheat of Neutralization = 316.50 J / 0.101moles                                                         = - 3133.746 J        

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