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?

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.

To determine the theoretical yield of precipitate for the double displacement reaction, we need to calculate the limiting reactant and use stoichiometry.

First, let's write the balanced chemical equation for the double displacement reaction between calcium nitrate (Ca(NO3)2) and sodium phosphate (Na3PO4):

3Ca(NO3)2 + 2Na3PO4 → Ca3(PO4)2 + 6NaNO3

From the balanced equation, we can see that the mole ratio between calcium nitrate and calcium phosphate is 3:1.

Next, let's calculate the number of moles of each reactant used:

Moles of calcium nitrate = volume (in L) × molarity = 0.025 L × 1.0 mol/L = 0.025 mol
Moles of sodium phosphate = volume (in L) × molarity = 0.015 L × 1.5 mol/L = 0.0225 mol

Since the mole ratio between calcium nitrate and calcium phosphate is 3:1, to find the limiting reactant, we need to compare the number of moles of calcium nitrate and sodium phosphate.

Since the moles of calcium nitrate (0.025 mol) are greater than the moles of sodium phosphate (0.0225 mol), calcium nitrate is not the limiting reactant. Therefore, sodium phosphate is the limiting reactant.

Now we can use the stoichiometry from the balanced equation to determine the theoretical yield of precipitate (calcium phosphate). According to the balanced equation, for every 2 moles of sodium phosphate, we get 1 mole of calcium phosphate.

The limiting reactant is sodium phosphate, with a mole quantity of 0.0225 mol. So, the theoretical yield of calcium phosphate (precipitate) is:

Theoretical yield = (0.0225 mol sodium phosphate) × (1 mol calcium phosphate / 2 mol sodium phosphate) = 0.01125 mol calcium phosphate

Therefore, the theoretical yield of precipitate (calcium phosphate) for the given reaction is 0.01125 moles.