Question

A student obtains 15.0 mL of 0.65 M phosphate solution and reacts it with excess ammonium and magnesium (Mg2+) under basic conditions. Upon reaction completion, she isolates and dries the product, and finds its mass to be 1.9426 g. Calculate the % yield

Answer 1

Answer on Question #55367 – Chemistry – General chemistry

Question:

A student obtains 15.0 mL of 0.65 M phosphate solution and reacts it with excess ammonium and magnesium (Mg2+) under basic conditions. Upon reaction completion, she isolates and dries the product, and finds its mass to be 1.9426 g. Calculate the % yield.

Solution:

$$\mathrm{PO_4^{3-}} + 3\mathrm{Mg^{2+}} = \mathrm{Mg_3(PO_4)_2\downarrow}$$

Ammonia doesn't precipitate at these conditions.

$$\mathrm{n(PO_4^{3-})} = \mathrm{c \times V} = 0.015 \times 0.65 = 0.00975\ (\mathrm{mol})\ (\text{or}\ 9.75\ \mathrm{mmol})$$ $$\mathrm{M(Mg_3(PO_4)_2)} = \mathrm{Ar(Mg) \times 3 + Ar(P) \times 2 + Ar(O) \times 8} = 24 \times 3 + 31 \times 2 + 16 \times 8 = 72 + 62 + 128 = 262$$ $$\mathrm{n(Mg_3(PO_4)_2)} = 0.00975\ (\mathrm{mol})$$ $$\mathrm{m(Mg_3(PO_4)_2)_T} = \mathrm{n(Mg_3(PO_4)_2) \times M(Mg_3(PO_4)_2)} = 0.00975 \times 262 = 2.5545\ (\mathrm{g})$$ $$\mathrm{Yield} = \mathrm{m_p/m_T} \times 100\% = 1.9426 / 2.5545 \times 100\% = 76.05\%$$

Answer: Yield 76.05 %

www.AssignmentExpert.com