Answer:
You will need 46.6 grams of iron(II) chloride and 173 mL of water.
Explanation:
x = mass of FeCl₂, y = mass of water
Assuming the percentage is by weight, we can write two equations:
x / (x + y) = 0.212
x + y = 220
Solving the system of equations:
x / 220 = 0.212
x ≈ 46.6
y ≈ 173
You will need 46.6 grams of iron(II) chloride and 173 mL of water.