Question

Copper metal reacts with a solution of silver nitrate, AgNO3, to produce copper (II) nitrate and silver metal. In carrying out this reaction, a piece of copper wire was immersed in a solution of silver nitrate until the reaction stopped. The original mass of the copper wire was 2.36 grams. After the reaction stopped, the mass of the wire was 1.03 grams. What mass of silver was produced?

Answer 1

Despite the long text, this is another stoichiometry question, I will follow the same step by step I have explained in our previous session

1. Balancing the equation:

Cu + 2 AgNO3 -> Cu(NO3)2 + 2 Ag, now the reaction is properly balanced

2. If we had 2.36 grams of Cu and only 1.03 grams are left, this means that we had copper in excess, and only 1.33 grams actually reacted, using the molar mass of Cu, 63.55 g/mol, we can find the number of moles:

63.55g = 1 mol

1.33g = x moles

x = 0.02 moles of Cu

3. Finding the molar ratio, which we can find by looking at the reaction, between Cu and Ag, the molar ratio is 1:2, which means 1 mol of Copper to produce 2 moles of silver, therefore if we have 0.02 moles of Cu, we will end up with 2 times this value, which is 0.04 moles of Ag

4. Now we have the number of moles of Ag, and its molar mass is, 107.87g/mol

107.87g = 1 mol

x grams = 0.04 moles

x = 4.31 grams of Silver will be produced from 1.33 grams of Copper