When chlorine and hydrogen interacted, 10.08 liters of hydrogen chloride were formed.

When chlorine and hydrogen interacted, 10.08 liters of hydrogen chloride were formed. Calculate the masses and amounts of substances that have reacted.

Let’s write down the solution, compose the reaction equation:
Cl2 + H2 = 2HCl – reaction of the compound, hydrogen chloride was obtained;
Let’s make calculations using the formulas of substances:
M (Cl2) = 2 * 35.5 = 71 g / mol;
M (HCl) = 1 + 35.5 = 36.5 g / mol;
Determine the amount of moles of hydrogen chloride:
1 mol of gas at n. y – 22.4 liters.
X mol (HCl) – 10.08 L from here, X mol (HCl) = 1 * 10.08 / 22.4 = 0.45 mol;
X mol (Cl2) – 0.45 mol (HCl);
-1 mol – 2 mol hence, X mol (Cl2) = 1 * 0.45 / 2 = 0.225 mol;
Find the amount of chlorine:
V (Cl2) = 0.225 * 22.4 = 5.04 L;
Chlorine mass is equal to:
m (Cl2) = Y * M = 0.225 * 71 = 15.97 g.
Let’s make the proportion:
X mol (H2) – 0.45 mol (HCl);
-1 mol – 2 mol from here, X mol (H2) = 1 * 0.45 / 2 = 0.225 mol;
Find the volume of hydrogen:
V (H2) = 0.225 * 22.4 = 5.04 liters.
The mass of hydrogen is:
m (H2) = Y * M = 0.225 * 2 = 0.45 g.
Answer: the volume of chlorine is 5.04 liters, the mass of chlorine is 15.97 g. The volume of hydrogen is 5.04 liters, the mass of hydrogen is 0.45 g.