Balance the chemical equation given below, and determine the number of milliliters of 0.00300 m phosphoric acid required to neutralize 25.00 ml of 0.00150 m calcium hydroxide. ca(oh)2(aq) + h3po4(aq) → ca3(po4)2(s) + h2o(l)

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

8.33 ml of phosphoric acid

Explanation:

To balance the equation, first add a 3 before Ca(OH)2, to balance Ca. Then, add a 2 before H3PO4, to balance P. Now you have 12 H on the left, so you have to add a 6 before H2O, to balance H. Finally, check Oxygen to notice that it is already balanced (there are 14 atoms at both sides).

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

In 0.00150 M Ca(OH)2 there are 0.0015 mol in 1 liter (or 1000 ml). Then in 25 ml, there are:

0.0015 mol / x mol = 1000 ml / 25 ml

x = 0.0015*25/1000

x = 3.75*10^(-5) mol of Ca(OH)2

From the balanced equation we know that 3 mol of Ca(OH)2 reacts with 2 mol of H3PO4, then 3.75*10^(-5) mol of Ca(OH)2 reacts with:

3 mol of Ca(OH)2 / 3.75*10^(-5) mol of Ca(OH)2 = 2 mol of H3PO4 / x mol of H3PO4

x = 2*3.75*10^(-5)/3

x = 2.5*10^(-5) mol of H3PO4

In 0.0030 M H3PO4 there are 0.003 mol in 1 liter (or 1000 ml). Then, 2.5*10^(-5) mol of H3PO4:

0.003 mol / 2.5*10^(-5) mol = 1000 ml / x ml

x = 1000*2.5*10^(-5)/0.003

x = 8.33 ml of H3PO4

the answer is c i believe

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