Consider the following reaction carried out to prepare zinc phosphate in the lab.
3Zn(NO3)2 (aq) + 2NA3(PO)4 -> Zn3(PO4)2 (s) + 6NaNO (aq)
A technician adds excess sodium phosphate solution to 5.00 L of 0.100 M zinc nitrate solution.
A) what is the theoretical yield of solid zinc phosphate?
B) If the process typically has a 97.0% yield, what mass would the technician expect to obtain when the precipitate is fully purified?
lyssa
Stoichiometry Questions And Answers
skipper
3Zn(NO3)2 + 2Na3PO4 --> Zn3(PO4)2 +6NaNO3
5.000 L x 0.100 M = 0.500 mole of Zn(NO3)2
0.500 mole Zn(NO3)2 x (1/3) = 0.1667 mole Zn3(PO4)2
0.1667 mole x 320.70 g/mole = 53.45 g Zn3(PO4)2 theoretical yield
53.456 x 0.970 = 51.85 g is a 97.0% yield
Roger the Mole
3 Zn(NO3)2(aq) + 2 Na3(PO4)(aq) → Zn3(PO4)2(s) + 6 NaNO3(aq)
A)
(5.00 L) x (0.100 mol/L Zn(NO3)2) x (1 mol Zn3(PO4)2 / 3 mol Zn(NO3)2) x (386.0827 g Zn3(PO4)2/mol) =
64.3 g Zn3(PO4)2
B)
(97.0% of 64.3 g Zn3(PO4)2) = 62.4 g Zn3(PO4)2