Question

3fecl₂ + 2na₃po₄ → fe₃(po₄)₂ + 6nacl
if 23 g of iron(ii) chloride reacts with excess sodium phosphate, what is the theoretical yield of the sodium chloride?
? g nacl

Explanation:

Step1: Calculate molar mass of FeCl₂

Molar mass of Fe = 55.85 g/mol, Cl = 35.45 g/mol.
Molar mass of \( \text{FeCl}_2 = 55.85 + 2\times35.45 = 126.75 \, \text{g/mol} \)

Step2: Moles of FeCl₂

Moles = \( \frac{\text{mass}}{\text{molar mass}} = \frac{23 \, \text{g}}{126.75 \, \text{g/mol}} \approx 0.1815 \, \text{mol} \)

Step3: Mole ratio from balanced equation

From \( 3\text{FeCl}_2 + 2\text{Na}_3\text{PO}_4
ightarrow \text{Fe}_3(\text{PO}_4)_2 + 6\text{NaCl} \),
mole ratio \( \text{FeCl}_2 : \text{NaCl} = 3 : 6 = 1 : 2 \).
So moles of \( \text{NaCl} = 0.1815 \, \text{mol} \times 2 = 0.363 \, \text{mol} \)

Step4: Molar mass of NaCl

Molar mass of Na = 22.99 g/mol, Cl = 35.45 g/mol.
Molar mass of \( \text{NaCl} = 22.99 + 35.45 = 58.44 \, \text{g/mol} \)

Step5: Mass of NaCl (theoretical yield)

Mass = moles × molar mass = \( 0.363 \, \text{mol} \times 58.44 \, \text{g/mol} \approx 21.2 \, \text{g} \)

Answer:

\( \approx 21 \) (or more precisely \( \approx 21.2 \)) g NaCl