In the reaction of sodium phosphate with aluminum nitrate, 65.0 mL of 0.125 mol/L sodium phosphate is reacted with 35.4 mL of 0.255 mol/L aluminum nitrate. What is the mass of solid that will form?

Na3PO4 + Al(NO3)3 —> AlPO4 + 3NaNO3

Mmol of Na3PO4 = 65 mL X 0.125 mol/L = 8.125 mmol of Na3PO4

Mmol of Al(NO3)3 = 35.4 mL X 0.255 mol/L = 9.027 mmol of Al(NO3)3

Since mmol of NaPO4 < mmol of Al(NO3)3, so Na3PO4 is the limiting reagent.

Here, 8.125 mmol of Na3PO4 will stoichiometrically react with 8.125 mmol of AlPO4,

Hence, the remaining Al(NO3)3 = (9.027 mmol — 8.125 mmol) = 0.902 mmol of Al(NO3)3.

At the end of reaction, the mass of solid existed will be:

• 0.902 mmol of Al(NO3)3 = 0.902 mmol X 212.996 g/mol [molar mass of Al(NO3)3] = 192.122 mg or 0.19 g of Al(NO3)3.
• mmol AlPO4 produced = 8.125 mmol or 8.125 mmol X 121.953 g/mol = 990.87 mg or 0.99 g of AlPO4.

Hope that helps.