During the electrolysis of the HCL solution at the cathode, 2 g of hydrogen was released over a period of time.

During the electrolysis of the HCL solution at the cathode, 2 g of hydrogen was released over a period of time. How much chlorine was released at the anode during the same time?

Data: mn (mass of released hydrogen) = 2 g.
Constants: k1 (electrochemical equivalent of chlorine) = 36.7 * 10-8 kg / C; k2 (electrochemical equivalent of hydrogen) = 1.045 * 10-8 kg / C.
1) We transform the first Faraday’s law for the given conditions: m = k * I * t and m / k = I * t.
2) Let’s compose and solve the equality: mн / k2 = mcl / k1, whence mcl = mн * k1 / k2.
Calculation: mcl = 2 * 36.7 * 10-8 / (1.045 * 10-8) = 70.24 g.
Answer: 70.24 grams of chlorine was released at the anode.