The addition of hydroiodic acid to a silver nitrate solution precipitates silver

Question

The addition of hydroiodic acid to a silver nitrate solution precipitates silver iodide according to the reaction:

AgNO3(aq)+HI(aq)AgI(s)+HNO3(aq)

When 50.0 mL of 5.00×102M  AgNO3 is combined with 50.0 mL of 5.00×102M HI in a coffee-cup calorimeter, the temperature changes from 22.40 C to 22.91 C.

Part A

Calculate Hrxn for the reaction as written. Use 1.00 g/mL as the density of the solution and Cs=4.18J/(gC) as the specific heat capacity of the solution.

Express the energy to two significant figures and include the appropriate units.

Explanation / Answer

We can calculate Hrxn for the given equation by the formula H = m×s×T where m is the mass of the reactants, s is the specific heat of the product, and T is the change in temperature from the reaction.

We have specific heat (s) = 4.18J/(gC), T = 22.91 - 22.40 = 0.51 C and to calculate m (the mass of the reactants) we have to use given molar values of AgNO3 and HI.

Change the given molar values of reactant in gram by multiplying molar mass of AgNO3 and HI.

Molar mass of AgNO3 = 169.87 g/mol and molar mass of HI = 127.911 g/mol.

So mass of AgNO3 = 5.00×10-2 M × 169.87 g/mol = 8.49 g and mass of HI = 5.00×10-2M×127.911 g/mol = 6.38 g

Total mass of reactants = 8.49 g + 6.38 g = 14.87 g

Now put the all values in formula we get H = m×s×T = (14.87g) × (4.18J/gC) × (0.51 C) = 31.70 Joule

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