Question

Reaction 1: Cu(s) + AgNO3 (aq) = CuNO3 (aq) + Ag (s)

Reaction 2: Cu (s) +AgNO3 (aq) = Cu(NO3)2 (aq) + Ag (s)

Part 1: Write the balanced net ionic equations for both of the above reactions.

Assume that excess silver nitrate is available when answering the remaining questions.

Part 2: Calculate the theoretical grams of silver metal that could form from 2.568g of copper wire based upon reaction 1. (Be sure the reaction is properly balanced.)

Part 3: Calculate the theoretical grams of silver metal that could form from 2.568g of copper wire based upon reaction 2. (Be sure the reaction is properly balanced.)

Part 4: Calculate the percent yield of silver produced according to each reaction if a student isolates 8.021g of Ag once the entire 2.568g of copper wire has been consumed. Based upon these yields, which is the most likely reeaction to have occured? Explain your answer.

Answer 1

Reaction 1. Cu_(s) + AgNO_3(aq) -----------> CuNO_3(aq) + Ag_(s)

Reaction 2. Cu_(s) + 2AgNO_3(aq) -----------> Cu(NO₃)_2(aq) + 2Ag_(s)

Part 1. Reaction 1: Cu_(s) + Ag⁺_(aq) + NO₃^-(aq) -----------> Cu⁺_(aq) + NO₃^-_(aq) + Ag_(s)

Cu_(s) + Ag⁺_(aq) -----------> Cu⁺_(aq)​ + Ag_(s)

Reaction 2: Cu_(s) + 2Ag⁺_(aq) + 2NO₃^-(aq) -----------> Cu²⁺_(aq)​ + 2NO₃^-_(aq) + 2Ag_(s)

Cu_(s) + 2Ag⁺_(aq)-----------> Cu²⁺_(aq)​ + 2Ag_(s)

Part 2: I'll use the net equation:

Cu_(s) + Ag⁺_(aq) -----------> Cu⁺_(aq)​ + Ag_(s)

atomic weight of copper: 63.55 g/mol

atomic weight of silver: 107.88 g/mol

Calculate moles: 2.568 / 63.55 = 0.0404 moles

1 mole of copper produce 1 mole of silver so: 0.0404 moles of Cu = 0.0404 moles of Ag

Mass of Ag = 0.0404 * 107.88 = 4.3584 g of silver produced

Part 3. The number of moles of Cu is the same, but the moles of Ag are different:

1 mole of Cu produces 2 mole of Ag so, 0.0404 moles * 2 = 0.0808 moles of Ag.

Mass of Ag = 0.0808 * 107.88 = 8.7167 g of silver produced

Part 4. According to the 8.021 g of silver produced, the yield would be:

Reaction 1: % = (8.021 / 4.3584) * 100 = 184%

Reaction 2: % = (8.021 / 8.7167) * 100 = 92.01%

Based on the results, the reaction that most likely occured was the second one, basically because reaction 2 and the % yield means first that is closer to this result, and second, the first result is practically the double amount of the obtained, and the the second reaction we have a Cu:Ag relation of 1:2 so, the result would have to be close to this one.