A 3.18 gram sample of copper metal is added to 105.5 grams of silver nitrate sol

Question

A 3.18 gram sample of copper metal is added to 105.5 grams of silver nitrate solution in a calorimeter. All of the copper metal reacts to form copper (II) nitrate, and metallic silver is precipitated. The original solution was at 19.5°C. The final temperature (after the reaction) is 37.5°C. Assume that the specific heat of the solution and copper combined is 3.80 J/g-°C.

a. Assuming that no energy leaks from the calorimeter, how many kilojoules of heat are produced by this reaction? (heat does actually leave the calorimeter but we assume all of the heat liberated by the reaction goes to warm the solution)

b. What is the H of reaction, express in kilojoules per mole of copper metal reacted? Be sure that the sign of H is correct

Explanation / Answer

ANSWER:

Following reaction takes place in calorimeter

Cu + AgNO3 -----------> CuNO3 + Ag(s)

Heat produced = Mass X Heat capacityX Temperature difference (T)

T = Tf - Ti = 37.5 - 19.5 = 18oC

one mole Cu (63.56g) reacts with one mole of AgNO3 (169.87g)

63.56 g of Cu= 169.87g of AgNO3

3.18g of Cu will react with 169.87/63.56 X 3.18 = 8.5g

Therefore total mass of reactants that will undergo reaction = 3.18 + 8.5 = 11.7g

Heat produced = 11.7 X 3.80 X 18 = 800.28J = 0.8KJ

b) Calculation of H of reaction per mole of Cu reacted

3.18g of Cu produces 0.8KJ of heat

3.18g of Cu = 0.8KJ of heat

One mole of Cu (63.56g) = 0.8/3.18 X 63.56 = 15.98KJ

H = - 15.98KJ

Ngative sign indicates energy is released

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