Question

in a double displacement creation 15.84g of ammonium phosphate completely reacts with excess lead (iv) nitrate, determine the mass of the precipitate formed when these 2 solutions are mixed

Answer 1

4(NH₄)₃PO₄(aq) + 3Pb(NO3)₄(aq) → Pb₃(PO₄)₄(s) + 12NH₄NO₃(aq)

M((NH₄)₃PO₄) = 149.08 g/mol

n = m/M

n((NH₄)₃PO₄) $= \frac{15.84}{149.08} = 0.106 \;mol$

n(Pb₃(PO₄)₄) = 1/4n((NH₄)₃PO₄) $= 0.25 \times 0.106 = 0.0265 \;mol$

$m = n \times M$

M(Pb₃(PO₄)₄) = 1001.48 g/mol

m(Pb₃(PO₄)₄) $= 0.0265 \times 1001.48 = 26.54 \;g$

Answer: 26.54 g