The specific heat of the product is 4.06jg deg. c and the, Ask an Expert

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The heat neutralization was determined for a reaction of HCl and NH3 by mixing 50.0mL of 1.05 M HCl with 50mL of 1.00 M NH3 in a calorimeter, producing an increasing temp from 23.0 deg. C to 29.6 deg. C for the solution of 0.5 M NH4Cl which formed as the product. The specific heat of the product is 4.06J/g deg. C and the density is 1.01g/mL.

a. Calculate the heat of the reaction, q, and the enthalpy per mole.

b. Why is the enthalpy based on the moles of NH3 and not HCl?

c. Calculate the percent error in part (a) if the accepted Hrxn is -55.0 kj/mol

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Category:- Chemistry
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