What is the mass of a 10% solution of silver nitrate subjected to electrolysis if, as a result of the reaction, oxygen with a volume of 1.12 liters is obtained?
1. Let’s compose the equation of electrolysis of a solution of silver nitrate, taking into account that silver will be reduced at the cathode, and water will be oxidized at the anode:
(-) K: Ag + + e- → Ag0;
(+) A: 2H2O – 4 e- → O2 + 4H +;
4AgNO3 + 2H2O = 4Ag + 4HNO3 + O2 ↑;
2.Calculate the chemical amount of evolved oxygen:
n (O2) = V (O2): Vm = 1.12: 22.4 = 0.05 mol;
3. Determine the amount of silver nitrate:
n (AgNO3) = n (O2) * 4 = 0.05 * 4 = 0.2 mol;
4.calculate the mass of nitrate:
m (AgNO3) = n (AgNO3) * M (AgNO3);
M (AgNO3) = 108 + 14 + 3 * 16 = 170 g / mol;
m (AgNO3) = 0.2 * 170 = 34 g;
5. find the mass of the solution subjected to electrolysis:
m (solution) = m (AgNO3): w (AgNO3) = 34: 0.1 = 340 g.
Answer: 340 g.