Answer:
![\large \boxed{\text{339 g}}](https://img.qammunity.org/2020/formulas/chemistry/middle-school/vstd4kgbs51uszakkwpvcmbmm807b9qvoc.png)
Step-by-step explanation:
We will need a balanced chemical equation with masses and molar masses, so, let's gather all the information in one place.
MM: 63.55 107.87
Cu + 2AgNO₃ ⟶ Cu(NO₃)₂ + 2Ag
m/g: 100
(a) Moles of Cu
![\text{Moles of Cu} = \text{100 g Cu }* \frac{\text{1 mol Cu}}{\text{63.55 g Cu}}= \text{1.574 mol Cu}](https://img.qammunity.org/2020/formulas/chemistry/middle-school/x8em92820jgoi1hn8ov2tq7xuyvnuygqmz.png)
(b) Moles of Ag
![\text{Moles of Ag} = \text{1.574 mol Cu} * \frac{\text{2 mol Ag}}{\text{1 mol Cu}} = \text{3.147 mol Ag}](https://img.qammunity.org/2020/formulas/chemistry/middle-school/248opkx7osjftwqtgw4gx08av9vjw8fn7p.png)
(c) Mass of Ag
![\text{Mass of Ag} =\text{3.147 mol Ag} * \frac{\text{107.87 g Ag}}{\text{1 mol Ag}} = \textbf{339 g Ag}\\\\\text{The reaction produces $\large \boxed{\textbf{339 g}}$ of Ag}](https://img.qammunity.org/2020/formulas/chemistry/middle-school/jx938jsujkmjwbf16vwz0slbnjurisvpos.png)