Answer:
127 g of Ag are made in the reaction.
Step-by-step explanation:
Reaction is: 2 AgNO₃ + Cu → Cu(NO₃)₂ + 2 Ag
If we only have data from nitrate, we assume the Cu is in excess.
We determine the moles of silver nitrate:
Mass / Molar mass → 200 g / 169.87 g/mol = 1.18 moles
Ratio is 2:2. If 2 moles of silver nitrate produce 2 moles of Ag, then
1.18 moles of salt, will produce 1.18 moles of silver
We convert mass to moles:
1.18 mol . 107.87 g/mol = 127 g