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A 6.65 g sample of aluminum chloride was added to 50.02 g of water. The temperature...

A 6.65 g sample of aluminum chloride was added to 50.02 g of water. The temperature rose from 22.1 o C to 28.0 o C. a. Write a net ionic equation for the process. b. What is the H per mole of aluminum chloride? c. How many kJ of energy would be produced if 2.75 g of the solute were dissolved?

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Answer #1

The net ionic equation will be

b) Heat Released = mass * specifc heat of water * change in temperature

=> 50.02 * 4.184 J/gC * (28.0-22.1)

=> 1234.77 J

Heat released if used 2.75g solute = 2.75/6.65 * 1234.77 = 510.691 J = 0.510KJ

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