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
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