Which one of the following represents the theoretical yield, in moles, of precipitate for the double displacement reaction if 25.0 mL of 1.0 M aqueous calcium nitrate is added to 15.0 mL of 1.5 M sodium phosphate?

Answer 1

The reaction is:
3Ca(NO3)2 + 2Na3PO4 --> Ca3(PO4)2 + 6NaNO3

moles of Ca(NO3)2 = (0.025L)(1.0mol/L) = 0.0250 mol
moles of Na3PO4 = (0.0150L)(1.5 mol/L) = 0.0225 mol

The mole ratio of Na3PO4/Ca(NO3)2 is 2/3
BUT we are mixing the two reagents at the ratio of 0.0225 / 0.0250 = 0.9 or 9/10. The excess reagent is Na3PO4. The LIMITING reagent is Ca(NO3)2.

[0.0250 mol Ca(NO3)2][1 mol. Ca3(PO4)2) / 3 mol. Ca(NO3)2] = 8.33x10^-3 moles of precipitate, Ca3(PO4)2.

Convert the above to grams to get the theoretical yield.