Balance the chemical equation given below, and determine the number of milliliters of 0.0300 m phosphoric acid required to neutralize 25.00 ml of 0.0150 m calcium hydroxide.
ca(oh)2(aq) + h3po4(aq) _ ca3(po4)2(s) + h2o(l)

1) Balanced equation:

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

2) Calculations

Moles

H3 PO4: ?
Ca[OH]2: 0.025 L * 0.0150 mol/L = 0.000375 mol

Ratios: 3 mol Ca[OH]2 / 2 mol H3PO4 = 0.000375 mol Ca[OH]2 / x mol H3PO4

=> x = 0.000375 * 2 /3 = 0.00025 mol Ca[OH]2

Volume of Ca[OH]2 = # of moles / Molarity = 0.00025 mol / 0.0300 mol/L = 0.00833 L = 8.33 mL

8.33 mL

The 2nd one I believe
Answer- 2nd