Answer to Question #263089 in General Chemistry for junior

Question

If 25 g of iron (II) chloride reacts with 48 g of sodium phosphate, what is the limiting reagent? How much sodium chloride and iron (II) phosphate can be formed

Answer 1

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

Molar mass of FeCl2= 126.751

25/126.751= 0.197

Molar mass of Na3PO4= 163.94

48/163.94= 0.293

= 0.293×3=0.878

= 0.197× 2= 0.394

Molar mass of NaCl= 58.44

= 1.272×58.44= 74.34g of NaCl