2h3po4 + 3mg(oh)2 → mg3(po4)2 + 6h2o phosphoric acid, h3po4, is neutralized by magnesium hydroxide, mg(oh)2, according to the equation shown. how many moles of water will be produced from the neutralization of 0.24 mole of h3po4? 0.72 mol 0.48 mol 0.24 mol 1.44 mol

The volume of the base Mg(OH)2 used is 5.28 L

Explanation:

Step 1: Data given

Volume of phosphoric acid = 5.12 L

Concentration of phosphoric acid = 2.75 M

concentration of magnesium hydroxide = 4.00 M

Step 2: The balanced equation

2H3PO4 + 3Mg(OH)2 → Mg3(PO4)2 + 6H2O

Step 3: Calculate the volume of magnesium hydroxide needed

a*Cb*Vb = b*Ca*Va

⇒with a = the coefficient of H3PO4 = 2

⇒with Cb = the concentration of Mg(OH)2 = 4.00 M

⇒with Vb = the volume of Mg(OH)2 = TO BE DETERMINED

⇒with b = the coefficient of Mg(OH)2 = 3

⇒with Ca = the concentration of H3PO4 = 2.75 M

⇒with Va = the volume of H3PO4 = 5.12 L

2 * 4.00 * Vb = 3 * 2.75 * 5.12

Vb = (3*2.75 * 5.12) / (2*4.00)

Vb = 5.28 L

The volume of the base Mg(OH)2 used is 5.28 L

5.28 L

Explanation:

Step 1:

Data obtained from the question.

Volume of acid (Va) = 5.12L

Molarity of acid (Ma) = 2.75M

Molarity of base (Mb) = 4M

Volume of base (Vb) =.?

Step 2:

The balanced equation for the reaction

2H3PO4 + 3Mg(OH)2 → Mg3(PO4)2 + 6H2O

From the balanced equation above,

Mole ratio of the acid (nA) = 2

Mole ratio of the base (nB) = 3

Step 3:

Determination of the volume of the base.

This is illustrated below:

MaVa/MbVb = nA/nB

2.75 x 5.12 / 4 x Vb = 2/3

Cross multiply

4 x 2 x Vb = 2.75 x 5.12 x 3

Divide both side by 4 x 2

Vb = (2.75 x 5.12 x 3)/(4 x 2)

Vb = 5.28 L

Therefore, the volume of the base is 5.28 L

The balanced equation for the neutralisation reaction is as follows

2H₃PO₄ + 3Mg(OH)₂ --> Mg₃(PO₄)₂ + 6H₂O

stoichiometry of H₃PO₄ to H₂O is 2:6

number of H₃PO₄ moles reacted - 0.24 mol

if 2 mol of H₃PO₄ form 6 mol of H₂O

then 0.24 mol of H₃PO₄ forms - 6/2 x 0.24 = 0.72 mol of H₂O

therefore 0.72 mol of H₂O are formed