A mixture is prepared by adding 22.7 mL of 0.157 M Na3PO4 to 31.2 mL of 0.277 M

Question

A mixture is prepared by adding 22.7 mL of 0.157 M Na3PO4 to 31.2 mL of 0.277 M Ca(NO3)2. (a) What mass of Ca3(PO4)2 will be formed? (b) What will be the concentrations of each of the ions in the mixture after reaction? Put your answers with 3 significant digits. Ca2+ Na+ PO43 NO3-

Explanation / Answer

3Ca(NO3)2 + 2Na3PO4 ---->Ca3(PO4)2 + 6 Na(NO3) 22.7 ml of 0.157 = 0.003564 moles Na3(PO4) 31.2 ml of 0.277 = 0.277 x31.2/1000 = 0.00864 moles Ca(NO3)2 0.003564 moles Na3(PO4) reacts with (3/2) x 0.003564 = 0.0005346 moles Ca(NO3)2 to give (1/2) x 0.003564 = 0.00178 moles Ca3(PO4)2 mass of Ca3(PO4)2 = 0.00178 x 310.18 = 0.55 gms

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