Answer:
[tex]\large \boxed{\text{339 g}}[/tex]
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
[tex]\text{Moles of Cu} = \text{100 g Cu }\times \dfrac{\text{1 mol Cu}}{\text{63.55 g Cu}}= \text{1.574 mol Cu}[/tex]
(b) Moles of Ag
[tex]\text{Moles of Ag} = \text{1.574 mol Cu} \times \dfrac{\text{2 mol Ag}}{\text{1 mol Cu}} = \text{3.147 mol Ag}[/tex]
(c) Mass of Ag
[tex]\text{Mass of Ag} =\text{3.147 mol Ag} \times \dfrac{\text{107.87 g Ag}}{\text{1 mol Ag}} = \textbf{339 g Ag}\\\\\text{The reaction produces $\large \boxed{\textbf{339 g}}$ of Ag}[/tex]
