The sum of the masses of all products of electrolysis of a solution containing 1 mol of silver nitrate.

Given:
n (AgNO3) = 1 mol

To find:
m (products) -?

Decision:
1) 4AgNO3 + 2H2O = (electrolysis) => 4Ag ↓ + 4HNO3 + O2 ↑;
2) M (Ag) = 108 g / mol;
M (HNO3) = 63 g / mol;
M (O2) = 32 g / mol;
3) n (Ag) = n (AgNO3) = 1 mol;
4) m (Ag) = n (Ag) * M (Ag) = 1 * 108 = 108 g;
5) n (HNO3) = n (AgNO3) = 1 mol;
6) m (HNO3) = n (HNO3) * M (HNO3) = 1 * 63 = 63 g;
7) n (O2) = n (AgNO3) / 4 = 0.25 mol;
8) m (O2) = n (O2) * M (O2) = 0.25 * 32 = 8 g;
9) m (products) = m (Ag) + m (HNO3) + m (O2) = 108 + 63 + 8 = 179 g.

Answer: The mass of products is 179 g.