Question

Cu + 2AgNO3 → Cu(NO³)2 + 2Ag (use this reaction to answer the following questions)

In a particular reaction between copper metal and silver nitrate (excess amounts), 12.7 g Cu was present to react,
how much pure silver Ag (moles) was produced?

Answer 1

0.4 mol

Step-by-step explanation:

1. Balanced Equation:

Cu + 2AgNO₃ → Cu(NO₃)₂ + 2Ag

2. As given in the question, 12.7g of Cu was present to react, therefore, this is the limiting reagent, and the silver nitrate is in excess. (also given).

n(Cu) = mass present divided by molar mass = m/M = 12.7/63.55 = 0.19984 mol

3. Since Stoichiometry = 1 : 2 : 1 : 2, therefore n(Ag) = 2×n(Cu) = 2(0.19984) = 0.399685... ≈ 0.4 mol of pure silver Ag produced