3PbCl2 72 mmol + 2(NH4)350 mmol PO4 ⟶ Pb3(PO4)2 + 6NH4Cl

Question

\(\underset{\text{72 mmol}}{3PbCl_2} + \underset{\text{50 mmol}}{2(NH_4)_3 }PO_4 \longrightarrow Pb_3(PO_4)_2 + 6NH_4Cl\)

Find mili mole of Pb₃(PO₄)₂ produced.

Answer 1

\(\underset{\text{72 mmol}}{3PbCl_2} + \underset{\text{50 mmol}}{2(NH_4)_3 }PO_4 \longrightarrow Pb_3(PO_4)_2 + 6NH_4Cl\)

\(\frac{nPbCl_2}{3} = \frac{nPb_3(PO_4)_2}1\)

\({nPb_3(PO_4)_2} = \frac{72}3 = 24 \) mmol