How many liters of chlorine should be taken to obtain 32.5 g of iron (III) chloride? Fe + Cl2 = FeCl3

2Fe + 3Cl2 = 2FeCl3

Let us find the amount of ferric chloride substance by the formula v = m / Mr, where m is the mass of the substance, v is the amount of the substance, Mr is the molecular mass.

Mr (FeCl3) = 162.2 g / mol

v = 32.5 / 162.2 = 0.2 mol

The reaction equation shows that chlorine and ferric chloride are in a ratio of 3: 2. Therefore, the amount of chlorine substance is 0.2 * 3/2 = 0.3 mol.

The volume of chlorine is found by the formula V = v * 22.4, where V is the volume of the substance, 22.4 l / mol is the molar volume of gases.

V = 0.3 * 22.4 = 6.72 l

Answer: to obtain 32.5 g of ferric chloride, 6.72 liters of chlorine are needed.