11.0k views
1 vote
How many grams of iron(II) chloride are needed to produce 44.3 g iron(II) phosphate in the presence of excess sodium phosphate?

User Earizon
by
6.9k points

1 Answer

1 vote

Answer:

47.2 g

Step-by-step 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

User Kuvonchbek Yakubov
by
5.9k points