Question

You react 2.33g of Iron(III) chloride with 50.0mL 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? (include balenced equation)

Answer 1

FeCl3 + Na3PO4 ---> FePO4 + 3 NaCl (aq)

no of moles of FeCl3 = weight / molar mass

= 2.33 / 162.2

= 0.0144 moles

no of moles of sodium phosphate

Molarity M = no of moles/ volume in liters

0.5 = no of moles / 0.05

no of moles of sodium phosphate = 0.5 x 0.05 = 0.025 moles

here FeCl3 is limiting agent so

according to balanced equation

one mole of FeCl3 and one mole of sodium phosphate will react and will give one mole of iron (III) phosphate and 3 moles of NaCl

similarly

0.0144 moles will give 0.0144 moles of iron (III) phosphate and 3 x 0.0144 moles NaCl

= 0.0432 moles

no w weight of NaCl

no of moles = weight / molar mass

weight = 0.0432 x 58.44

= 2.5246 grams