Determine the mass of iron required to obtain 43.5 g of iron (III) chloride.

From 1 mole of iron, 1 mole of its chloride can be synthesized.

Let’s calculate the chemical amount of a substance in 43.5 grams of ferric chloride.

M FeCl3 = 56 + 35.5 x 3 = 162.5 grams / mol;

N FeCl3 = 43.5 / 162.5 = 0.268 mol;

The same amount of iron is needed.

Let’s calculate its weight.

M Fe = 56 grams / mol;

m Fe = 56 x 0.268 = 15 grams;