You react 2.33 g of iron (III) chloride with 50.0 mL of 0.500 M solution of sodium phosphate to form iron (III) phosphate and sodium chloride. How many grams of sodium chloride ought to
be produced?

2.33g of iron (iii) chloride

50.0 mL of 5.00 M of sodium phosphate

FeCl3 + Na3PO4 > Fe(PO4) + 3NaCl

mol = conc × vol = 0.5 × 50/1000 = 0.025 mol Na3PO4

from the equation:

1 mol of Na3PO4 reacts with 1 mol FeCl3 = 3 mol of NaCl

0.025 mol = x

x = 0.0025 × 3 = 0.075 mol NaCl

mass = 0.075 g × 59 g/mol = 4.425 g NaCl

i guessed all of this so i dont know i it is correct