What mass of iron Fe and what volume of chlorine Cl2 must react to form 48.75 g of iron (III) chloride?

1. Let’s write down the reaction equation:

2Fe + 3Cl2 = 2FeCl3.

2. Find the amount of ferric chloride:

n (FeCl3) = m (FeCl3) / M (FeCl3) = 48.75 g / 162.5 g / mol = 0.3 mol.

3. According to the reaction equation, we find the amount of iron and chlorine:

n (Fe) = n (FeCl3) = 0.3 mol.

m (Fe) = n (Fe) * M (Fe) = 0.3 mol * 56 g / mol = 16.8 g.

n (Cl2) = n (FeCl3) / 2 * 3 = 0.3 mol / 2 * 3 = 0.2 mol.

V (Cl2) = n (Cl2) * Vm = 0.2 mol * 22.4 L / mol = 4.48 L.

Answer: m (Fe) = 16.8 g; V (Cl2) = 4.48 l.