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.

c is the answer because when something is left after something evaporates that shows the chemical change.

it should be all of them because it asks if you included all of those in your answer

explanation: