Question

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?

Answer 1

Answer:

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