7 g of iron reacted with 18 g of chlorine. What is the mass of ferric chloride 3 formed?

Let’s write the reaction equation:

2Fe + 3Cl2 → 2FeCl3 (iron (III) chloride).

Let’s calculate the amount of iron substance:

n (Fe) = m (Fe) / M (Fe) = 7 g / 56 g / mol = 0.125 mol.

Let’s calculate the amount of chlorine substance:

n (Cl2) = m (Cl2) / M (Cl2) = 18 g / 71 g / mol = 0.254 mol.

For 0.125 mol of iron, 0.1875 mol of chlorine is required, which means chlorine is in excess. We are calculating the hardware.

Let’s calculate the amount of iron (III) chloride substance:

n (FeCl3) = n (Fe) = 0.125 mol.

We calculate the mass of iron (III) chloride:

m (FeCl3) = n (FeCl3) * M (FeCl3) = 0.125 mol * 162.5 g / mol = 20.3125 g.

Answer: 20.3125 grams.