Answer:
50.9 g of Ag
Step-by-step explanation:
Reaction is:
Cu (s) + 2AgNO₃(aq) → 2Ag (s) + Cu(NO₃)₂ (aq)
1 mol of Cu react to 2 moles of silver nitrate in order to produce 2 moles of solid silver and 1 mol of copper (II) nitrate.
We assume, the silver nitrate in excess.
We determine moles of copper → 15 g . 1mol / 63.55 g = 0.236 moles
Ratio is 1:2. 1 mol of copper produce 2 moles of silver
0.236 moles of Cu will produce (0.236 . 2) / 1 = 0.472 moles of Ag
We deterine mass of silver:
0.472 mol . 107.87 g /mol = 50.9 g