A 2.7 M solution of phosphoric acid (H3PO4) is to be reacted with a 7.5 M solution of sodium hydroxide to make sodium phosphate and water. How many L of the H3PO4 solution will react with 0.662 L of the NaOH solution
H3PO4 + 3NaOH ==> 3H2O + Na3PO4
mols NaOH = M x L = ?
Convert mols NaOH to mols H3PO4 using the coefficients in the balanced equation.
M H3PO4 = mols/L. You have M and mols, solve for L H3PO4.
To determine the volume of the H3PO4 solution that will react with 0.662 L of the NaOH solution, we need to use the concept of stoichiometry.
The balanced equation for the reaction between phosphoric acid (H3PO4) and sodium hydroxide (NaOH) is:
H3PO4 + 3NaOH -> Na3PO4 + 3H2O
From the balanced equation, we can see that 1 mole of H3PO4 reacts with 3 moles of NaOH. Therefore, the mole ratio between H3PO4 and NaOH is 1:3.
Now, let's calculate the number of moles of NaOH in 0.662 L of a 7.5 M (7.5 moles per liter) solution:
moles of NaOH = concentration of NaOH x volume of NaOH solution
moles of NaOH = 7.5 mol/L x 0.662 L
moles of NaOH = 4.965
Since the mole ratio between H3PO4 and NaOH is 1:3, we can say that 4.965 moles of NaOH will react with (4.965/3) moles of H3PO4.
Now, we need to find the volume of the H3PO4 solution that contains (4.965/3) moles. We can do this by using the concentration of the H3PO4 solution.
0.662 L of the NaOH solution is used, and we need to determine the volume of the H3PO4 solution that contains the same number of moles.
moles of H3PO4 = (4.965/3)
concentration of H3PO4 = 2.7 M (2.7 mol/L)
volume of H3PO4 solution = moles of H3PO4 / concentration of H3PO4
volume of H3PO4 solution = (4.965/3) / 2.7
volume of H3PO4 solution ≈ 0.617 L
Therefore, approximately 0.617 liters of the 2.7 M solution of phosphoric acid will react with 0.662 liters of the 7.5 M solution of sodium hydroxide.