Question

3fecl₂ + 2na₃po₄ → fe₃(po₄)₂ + 6nacl
if 23 g fecl₂ reacts with 41 g na₃po₄,
what is the limiting reactant?

1. fecl₂ 2. na₃po₄

enter the answer choice number.

Explanation:

Step1: Calculate moles of FeCl₂

Molar mass of \( \text{FeCl}_2 \) is \( 55.85 + 2\times35.45 = 126.75 \, \text{g/mol} \).
Moles of \( \text{FeCl}_2 = \frac{23 \, \text{g}}{126.75 \, \text{g/mol}} \approx 0.1815 \, \text{mol} \).

Step2: Calculate moles of Na₃PO₄

Molar mass of \( \text{Na}_3\text{PO}_4 \) is \( 3\times22.99 + 30.97 + 4\times16.00 = 163.94 \, \text{g/mol} \).
Moles of \( \text{Na}_3\text{PO}_4 = \frac{41 \, \text{g}}{163.94 \, \text{g/mol}} \approx 0.2499 \, \text{mol} \).

Step3: Compare mole ratios

From the reaction: \( 3 \, \text{FeCl}_2 : 2 \, \text{Na}_3\text{PO}_4 \) (mole ratio \( 3:2 \)).
For \( \text{FeCl}_2 \), required \( \text{Na}_3\text{PO}_4 \) moles: \( \frac{2}{3} \times 0.1815 \approx 0.121 \, \text{mol} \).
Available \( \text{Na}_3\text{PO}_4 \) is \( 0.2499 \, \text{mol} \) (excess).
For \( \text{Na}_3\text{PO}_4 \), required \( \text{FeCl}_2 \) moles: \( \frac{3}{2} \times 0.2499 \approx 0.3749 \, \text{mol} \).
Available \( \text{FeCl}_2 \) is \( 0.1815 \, \text{mol} \) (insufficient).
Thus, \( \text{FeCl}_2 \) is limiting.

Answer:

1