A mixture is prepared by adding 28.1 mL of 0.176 M Na3PO4 to 33.6 mL of 0.236 M

Question

A mixture is prepared by adding 28.1 mL of 0.176 M Na3PO4 to 33.6 mL of 0.236 M Ca(NO3)2.



(a) What mass of Ca3(PO4)2 will be formed?
.767 g Ca3(PO4)2


(b) What will be the concentrations of each of the ions in the mixture after reaction? Put your answers with 3 significant digits.

Explanation / Answer

2Na3PO4 + 3Ca(NO3)2 --> Ca3(PO4)2 + 6NaNO3 0.0255L x 0.173M = 0.00441moles Na3PO4. 0.0132moles Na+, 0.00441moles PO4 3- 0.0325L x 0.258M = 0.00838moles Ca(NO3)2. 0.00838moles Ca2+, 0.01676moles NO3- 0.00441moles Na3PO4 x (3 Ca(NO3)2 / 2 Na3PO4) = 0.00662moles Ca(NO3)2 required we have 0.00838moles, Na3PO4 = limiting reactant

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