Calculate the mass, the amount of the substance that reacted iron with chlorine if 0.7 mol of FeCl3 was formed.

Data: νCl – the amount of the substance of the obtained chlorine (νCl = 0.7 mol).

Const: MFe – molar mass of iron (MFe = 55.845 g / mol); MCl2 is the molar mass of chlorine (MCl2 = 70.906 g / mol).

1) The reaction of combining the desired mass of iron with chlorine: 2FeCl3 = 2Fe + 3Cl2.

2) The amount of substance reacted Fe: 2νFe = 2νFeCl3 and νFe = νFeCl3 = 0.7 mol.

The amount of substance reacted Cl2: 3νCl2 = 2νFeCl3 and νCl2 = 2νFeCl3 / 3 = 3 * 0.7 / 2 = 1.05 mol.

3) Fe mass: mFe = νFe * MFe = 0.7 * 55.845 ≈ 39.1 g.

Cl2 mass: mСl2 = νCl2 * МСl2 = 1.05 * 70.906 ≈ 74.45 g.

Answer: 39.1 g of iron and 74.45 g of chlorine should have entered into the reaction.