What is the maximum amount of Na3PO4 that is formed from the reaction of 25.00 mL of 0.1050 M NaOH and 15.00 mL of 0.08650 M H3PO4?

Question

NaOH + H3PO4 → Na3PO4 + H2O

For the second part,

What is the molarity of Na3PO4 in the solution prepared in the previous problem?

Answer 1

First, we need to determine the limiting reactant of the reaction between NaOH and H3PO4. To do this, we can use the balanced chemical equation and calculate the number of moles of each reactant:

NaOH + H3PO4 → Na3PO4 + H2O

Moles of NaOH = (0.1050 M) × (25.00 mL/1000 mL) = 2.625 × 10^-3 mol

Moles of H3PO4 = (0.08650 M) × (15.00 mL/1000 mL) = 1.298 × 10^-3 mol

Based on the stoichiometry of the balanced equation, we can see that 1 mole of NaOH reacts with 1 mole of H3PO4 to produce 1 mole of Na3PO4. Therefore, since we have less moles of H3PO4, it is the limiting reactant. The amount of Na3PO4 formed can be calculated by multiplying the number of moles of H3PO4 by the mole ratio of Na3PO4 to H3PO4:

1 mol Na3PO4/1 mol H3PO4 × 1.298 × 10^-3 mol H3PO4 = 1.298 × 10^-3 mol Na3PO4

Now, to calculate the molarity of Na3PO4 in the solution, we need to know the total volume of the solution. This can be calculated by adding the volumes of NaOH and H3PO4 used:

Total volume = 25.00 mL + 15.00 mL = 40.00 mL = 0.04000 L

The molarity of Na3PO4 can then be calculated by dividing the number of moles of Na3PO4 by the total volume of the solution:

Molarity of Na3PO4 = 1.298 × 10^-3 mol/0.04000 L = 0.03245 M

Therefore, the maximum amount of Na3PO4 that can be formed from the given reactants is 1.298 × 10^-3 mol, and the molarity of Na3PO4 in the resulting solution is 0.03245 M. upvote my answer