Question

3fecl₂ + 2na₃po₄ → fe₃(po₄)₂ + 6nacl
you found that fecl₂ is the limiting reactant. what mass of fe₃(po₄)₂ forms from 23 g fecl₂ during the reaction?
? g fe₃(po₄)₂

Explanation:

Step1: Calculate moles of \( \text{FeCl}_2 \)

Molar mass of \( \text{FeCl}_2 \): \( \text{Fe} = 55.85 \, \text{g/mol} \), \( \text{Cl} = 35.45 \, \text{g/mol} \).
Molar mass \( = 55.85 + 2 \times 35.45 = 126.75 \, \text{g/mol} \).
Moles of \( \text{FeCl}_2 = \frac{\text{mass}}{\text{molar mass}} = \frac{23 \, \text{g}}{126.75 \, \text{g/mol}} \approx 0.1815 \, \text{mol} \).

Step2: Use stoichiometry to find moles of \( \text{Fe}_3(\text{PO}_4)_2 \)

From the reaction: \( 3 \, \text{mol} \, \text{FeCl}_2
ightarrow 1 \, \text{mol} \, \text{Fe}_3(\text{PO}_4)_2 \).
Moles of \( \text{Fe}_3(\text{PO}_4)_2 = \frac{1}{3} \times \text{moles of } \text{FeCl}_2 = \frac{1}{3} \times 0.1815 \, \text{mol} \approx 0.0605 \, \text{mol} \).

Step3: Calculate molar mass of \( \text{Fe}_3(\text{PO}_4)_2 \)

\( \text{Fe} = 55.85 \, \text{g/mol} \), \( \text{P} = 30.97 \, \text{g/mol} \), \( \text{O} = 16.00 \, \text{g/mol} \).
Molar mass \( = 3 \times 55.85 + 2 \times (30.97 + 4 \times 16.00) \)
\( = 167.55 + 2 \times (30.97 + 64.00) \)
\( = 167.55 + 2 \times 94.97 \)
\( = 167.55 + 189.94 = 357.49 \, \text{g/mol} \).

Step4: Calculate mass of \( \text{Fe}_3(\text{PO}_4)_2 \)

Mass \( = \text{moles} \times \text{molar mass} = 0.0605 \, \text{mol} \times 357.49 \, \text{g/mol} \approx 21.6 \, \text{g} \).

Answer:

\( \approx 22 \) (or more precisely \( \approx 21.6 \))