How much iron can react with 67.2 liters of chlorine?

Let’s execute the solution:

In accordance with the condition of the problem, we write down the process:
Xr -? V = 67.2 l

2Fe + 3Cl2 = 2FeCl3 – compounds, ferric chloride was obtained;

Let’s make the calculations:
M (Fe) = 55.8 g / mol;

M (Cl2) = 71 g / mol.

Proportions:
1 mol of gas at normal level – 22.4 liters;

X mol (Cl2) – 67.2 liters from here, X mol (Cl2) = 1 * 67.2 / 22.4 = 3 mol;

X mol (Fe) – 3 mol (Cl2);

-2 mol -3 mol from here, X mol (Fe) = 2 * 3/3 = 2 mol.

Find the mass of the metal:
m (Fe) = Y * M = 2 * 55.8 = 111.6 g

Answer: you need iron weighing 111.6 g




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