A mixture is prepared by adding 26.3 mL of 0.176 M Na3PO4 to 31.4 mL of 0.269 M

Question

A mixture is prepared by adding 26.3 mL of 0.176 M Na3PO4 to 31.4 mL of 0.269 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+ M
Na+ M
PO43- M
NO3

Explanation / Answer

26.3 mL of 0.176 M Na3PO4 to 31.4 mL of 0.269 M Ca(NO3)2. 2Na3PO4 + 3Ca(NO3)2 ---> Ca3(PO4)2+ 6NaNo3 so 26.3 mL of 0.176 M Na3PO4=4.63mmol 31.4 mL of 0.269 M Ca(NO3)2. =8.45mol a) moles of Ca3(PO4)2 will be= 4.63*0.5mmol=2.315 mmol mass=2.315*10^-3*310=0.717gm b)total volume=57.7ml mass of Ca(NO3)2.left will be= 1.5mmol Ca2+ = 0.146M Na+ =0.24 M PO43- = 0.08M NO3 =0.77M

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