Question

How many grams of Cu are needed to react with 105.2g of Ag

Answer 1

$2Ag^+ + Cu = Cu(2+) + 2 Ag$

$n(Ag) = 105.2/107.87 mol = 0.975 mol$

$n(Cu) = n(Ag)/2 = 0.4875 mol => m(Cu) = 0.4875*63.55 = 30.98 g$