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.

a non redox reaction occurs when there is not a transfer of electrons. for redox reactions, which are reduction-oxidation reactions, a change in the oxidation number occurs. elements that are reduced gain electrons, and elements that are oxidized lose electrons.

explanation:

each  column  of the  table  is called a group. elements in a group share similar properties. groups are  read  from  top  to  bottom. metal. metalloid. nonmetal. solid. liquid. gas. fe. hg. o.  2. 3. 4. 5. 6. 7. 8. 9.  2. beryllium. 9.012. 4.  sodium. 22.990.  11. 3.  magnesium. 24.305.  12.  potassium. 39.098.  19. 4.  calcium. 40.078.  20. i guess  i