Question

A reaction that occurs when 23 grams of iron (II) chloride reacts with sodium phosphate forming iron (II) phosphate and sodium chloride. what is the limiting reagent? How much sodium chloride can be formed?

Answer 1

Balanced Equation

3FeCl₂+2Na₃PO₄$\implies$ Fe₃(PO₄)₂+6NaCl

Limiting reagent is FeCl₂

Moles of FeCl₂ =$\frac{23g}{126.751g/mol}$

=0.1815moles

Mole ratio of:

FeCl₂:NaCl

3 : 6

If 3=0.1815moles

$\therefore$ 6=?

$\frac{6×0.1815}{3}$ $=0.363moles$

Molar mass of NaCl is 58.44g/mol

But Mass= 0.363moles×58.44g/mol

=21g

Mass Of Sodium Chloride formed =21g