How many grams of iron (II) chloride are needed to produce 44.3 g iron (II) phosphate in the presence of excess sodium phosphate?

47.2 g

Explanation:

Let's consider the following double displacement reaction.

3 FeCl₂ + 2 Na₃PO₄ → Fe₃ (PO₄) ₂ + 6 NaCl

The molar mass of Fe₃ (PO₄) ₂ is 357.48 g/mol. The moles corresponding to 44.3 g are:

44.3 g * (1 mol / 357.48 g) = 0.124 mol

The molar ratio of Fe₃ (PO₄) ₂ to FeCl₂ is 1:3. The moles of FeCl₂ are:

3 * 0.124 mol = 0.372 mol

The molar mass of FeCl₂ is 126.75 g/mol. The mass of FeCl₂ is:

0.372 mol * (126.75 g/mol) = 47.2 g