How many grams of sodium chloride can be formed when 25 grams of iron(II)....

How many grams of sodium chloride can be formed when 25 grams of iron(II) chloride react with 50. grams of sodium phosphate according to the following balance chemical equation?

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

Limiting reagent question.

Step 1 balanced equation. 3 FeCl2 + 2 Na3PO4 → Fe3(PO4)2 + 6 NaCl

Step 2. change all masses into moles.

moles FeCl2 = 25 g / 126.75 g/mol = 0.197 moles

moles Na3PO4 = 50 g / 163.94 g/mol = 0.305 moles

Step 3. Check the mole ratio between the two reactants. It is 3 to 2. So for every mole of FeCl2 you have you will need 2/3 that many moles of Na3PO4. So you have 0.197 moles x 2/3 = 0.131 moles of Na3PO4 needed to use up ALL the FeCl2. You have way more than that. So the FeCl2 is the limiting reagent and Na3PO4 is in excess.

Step 4. Mole ratio between FeCl2 and NaCl is 3 to 6. Or you get twice the number of moles of NaCl as you have of FeCl2

So moles of NaCl will be 0.197 x 2 = 0.394 moles

Step 5. Convert the moles of NaCl into a mass

mass = moles x molar mass.

You do the last calculation.