calculate the volume of 1.07M phosphoric acid solution necessary to react with 22.9 ml of 0.150M Mg(OH)

2H3PO4 +3Mg(OH)2 ==> Mg(PO4)2 + 6H20

I've corrected the equation.

2H3PO4 +3Mg(OH)2 ==> Mg3(PO4)2 + 6H20

mols Mg(OH)2 = M x L = ?
Convert mols Mg(OH)2 to mols H3PO4 using the coefficients in the balanced equation.
Then mols H3PO4 = M H3PO4 x L H3PO4 You know mols and M, solve for L.

To calculate the volume of the phosphoric acid solution needed to react with a given volume of magnesium hydroxide solution, we will first determine the number of moles of each reactant.

Let's start by calculating the moles of magnesium hydroxide using the given concentration and volume:

Moles of Mg(OH)2 = Concentration * Volume = 0.150 M * 0.0229 L = 0.003435 moles

Next, we will use the balanced chemical equation to determine the mole ratio between magnesium hydroxide and phosphoric acid. According to the equation:

2H3PO4 + 3Mg(OH)2 ==> Mg(PO4)2 + 6H2O

The mole ratio between Mg(OH)2 and H3PO4 is 3:2. Therefore, the moles of phosphoric acid will be:

Moles of H3PO4 = (3/2) * Moles of Mg(OH)2 = (3/2) * 0.003435 = 0.0051525 moles

Finally, we can calculate the volume of the phosphoric acid solution using its concentration and the moles we just calculated:

Volume of H3PO4 = Moles / Concentration = 0.0051525 moles / 1.07 M = 0.004822 L = 4.822 mL

Therefore, approximately 4.822 mL of the 1.07 M phosphoric acid solution is necessary to react with 22.9 mL of the 0.150 M magnesium hydroxide solution.

To calculate the volume of the phosphoric acid solution needed to react with the given volume of magnesium hydroxide solution, we can use the concept of stoichiometry and the balanced chemical equation.

Let's break down the question and the given information:

1. Concentration of phosphoric acid solution: 1.07 M.
2. Volume of magnesium hydroxide solution: 22.9 ml.
3. Concentration of magnesium hydroxide solution: 0.150 M.

Now, let's determine the stoichiometry of the balanced chemical equation:
From the balanced equation: 2H3PO4 + 3Mg(OH)2 → Mg3(PO4)2 + 6H2O

We can observe that 2 moles of phosphoric acid (H3PO4) react with 3 moles of magnesium hydroxide (Mg(OH)2) to form 1 mole of magnesium phosphate (Mg3(PO4)2).

Now, we can start solving the problem:

Step 1: Calculate the number of moles of magnesium hydroxide (Mg(OH)2) using its concentration and volume.
Moles of Mg(OH)2 = Molarity × Volume (in liters)
Moles of Mg(OH)2 = 0.150 M × 0.0229 L (since the volume is given in ml, we convert it to liters by dividing by 1000)
Moles of Mg(OH)2 = 0.003435 moles

Step 2: Determine the stoichiometric ratio between magnesium hydroxide (Mg(OH)2) and phosphoric acid (H3PO4).
From the balanced equation, we know that 2 moles of H3PO4 react with 3 moles of Mg(OH)2.
Therefore, there is a 2:3 ratio between H3PO4 and Mg(OH)2.

Step 3: Calculate the number of moles of phosphoric acid (H3PO4) required to react with the given moles of Mg(OH)2.
Moles of H3PO4 = (Moles of Mg(OH)2 × Stoichiometric coefficient of H3PO4) / Stoichiometric coefficient of Mg(OH)2
Moles of H3PO4 = (0.003435 moles × 2) / 3
Moles of H3PO4 = 0.00229 moles

Step 4: Calculate the volume of the phosphoric acid solution using its concentration and the number of moles required.
Volume of H3PO4 = Moles of H3PO4 / Molarity
Volume of H3PO4 = 0.00229 moles / 1.07 M
Volume of H3PO4 = 0.002139 L (or 2.139 ml)

Therefore, the volume of the 1.07 M phosphoric acid solution required to react with 22.9 ml of 0.150 M Mg(OH)2 is approximately 2.139 ml.