An aqueous of h3po4 was prepared by dissolving 8.85 g in enough water to make 350.0 ml solution. also, an aqueous solution of ca(oh)2 was prepared by dissolving 15.76 g of ca(oh)2 in enough water to make 550 ml solution. calculate the volume of h3po4 required to neutralize 25.0 ml of ca(oh)2

25.0 mL

Explanation:

1. Gather the information in one place.

MM:      98.00         74.09

           2H3PO4 + 3Ca(OH)2 → Ca3(PO4)2 + 6H2O

m/g:       8.85           15.76

V/mL:   350.0           550

2. Moles of H3PO4

n = 8.85 g × (1 mol/98.00 g) = 0.09031 mol H3PO4

3. Moles of Ca(OH)2

n = 15.76 g × (1 mol/74.09 g) = 0.2126 mol Ca(OH)2

4. Moles of Ca(OH)2 in 25.0 mL Solution

n = 0.2126 mol × (25.0 mL/550 mL) = 0.009 663 mol Ca(OH)2

5. Moles of H3PO4 needed

From the balanced equation, the molar ratio is 2 mol H3PO4: 3 mol Ca(OH)2

n = 0.009 663 mol Ca(OH)2 × (2 mol H3PO4/3 mol Ca(OH)2)

= 0.006 442 mol H3PO4

6. Volume of H3PO4

V = 0.006 442 mol ×( 350.0 mL/0.09031 mol) = 25.0 mL H3PO4

It will take 25.0 mL of the H3PO4 solution to neutralize 25.0 mL of the Ca(OH)2 solution.

i think its 5049.

i hope this !