Chemistry Question PLEASEEE help

Question

A student decides to find the molar enthalpy for the reaction between phosphoric acid and calcium hydroxide. They have a 0.250 M solution of Ca(OH)₂, which they will use; they can only find a 2.1 M stock solution of H₃PO₄. If they want 285 mL of a 0.31 solution of H₃PO₄, how many mL of the stock solution should they use to prepare their solution?

The same student decides to simply prepare a 0.28 M solution of H3PO4. How many mL of this solution are required to react stoichiometrically with 125 mL of their 0.250 M Ca(OH)2 solution?

3Ca(OH)2 (aq) + 2H3PO4 (aq) ---> Ca3(PO4)2 (s) + 6H2O (l)

Give your answer to the nearest tenth of a mL (e.g.: 52.8 mL)

Answer 1

First part:

Use V1M1 = V2M2 for such a dilution problem.

V1 = initial volume = ?

M1 = initial molarity = 2.1 M

V2 = final volume = 285 ml

M2 = final molarity = 0.31 M

(V1)(2.1)( = (285)(0.31)

V1 = 40.7 mls of 2.1 M H₃PO₄ are needed

Second part:

Using 0.28 M H₃PO₄, find mls needed to react with 125 ml of 0.250 M Ca(OH)₂.

The balanced equation for this reaction is: 3Ca(OH)₂ (aq) + 2H₃PO₄ (aq) ---> Ca₃(PO₄)₂ (s) + 6H₂O (l)

First we'll find the moles of Ca(OH)2 present:

125 ml x 1 L/1000 ml x 0.250 mol/L = 0.03125 mols Ca(OH)₂ present

From the balanced equation, we now find moles of H₃PO₄ needed:

0.03125 mols Ca(OH)₂ x 2 mols H₃PO₄ / 3 mols Ca(OH)₂ = 0.0208 mols H₃PO₄ needed

Finally, we find volume (mls) of H₃PO₄ needed:

0.0208 mols H₃PO₄ x 1 L / 0.28 mols = 0.0743 L x 1000 ml/L = 74.3 mls H₃PO₄ needed