How many grams of berthollet’s salt would be obtained by passing 11.2 liters of chlorine through

How many grams of berthollet’s salt would be obtained by passing 11.2 liters of chlorine through a hot potassium hydroxide solution?

To solve, we write down the reaction equation:
6KOH + 3Cl2 = KClO3 + 5KCl + 3H2O – redox reaction, Berthollet’s salt was formed:
Let us determine the molar mass of the berthollet salt:
M (KClO3) = 39.1 + 35.5 + 16 * 3 = 112.6 g / mol;
Let us calculate the number of moles of Cl2:
1 mol of gas at n. y occupies a volume of 22.4 liters;
X mol (Cl2) – 11.2 liters. hence, X mol (Cl2) = 1 * 11.2 / 22.4 = 0.5 mol;
Let’s make a proportion according to the reaction equation:
0.5 mol (Cl2) – X mol (KClO3);
-3 mol – 1 mol from here, X mol (KClO3) = 0.5 * 1/3 = 0.16 mol;
Let us determine the mass of the berthollet salt:
m (KClO3) = Y * M = 0.16 * 112, 6 = 18 g.
Answer: in the course of the reaction, Berthollet’s salt with a mass of 18 g was released.