Calculate the moles of acetic acid used in each trial and record in Data Table 3.Volume acetic acid in Liters = (Mass/Density)/1000Moles = Volume ∗ Concentration Calculate the moles of magnesium hydroxide (Mg(OH)2) used in each trial and record in Data Table 3.Moles Mg(OH2) = Moles acetic acid
Calculate the neutralization capacity of each trial and record in Data Table. Neutralization Capacity = Moles Mg(OH)2 / Mass of Milk of Magnesium
Data Table Trial 1 Trial 2
Mass of milk of magnesia 2.5g 2.5g
Density of milk of magnesia 1.14 g/mL 1.14 g/mL
Volume of acetic acid, initial 10mL 10mL
Volume of acetic acid, final 2.2mL 1.8mL
Volume of acetic acid, total 7.8mL 8.2mL
Concentration of acetic acid 0.88 M 0.88 M
Moles of acetic acid
Moles of Mg(OH)2
Moles Mg(OH)2 / g milk of magnesia

Trial 1: Moles acetic acid = 0.00686 moles;

Moles of Mg(OH)₂ = 0.00343 moles

Neutralization capacity = 0.00137 mol/g

Trial 2: Moles acetic acid = 0.00722 moles

Moles of Mg(OH)₂ = 0.00361 moles

Neutralization capacity = 0.00144 mol/g

Explanation:

Equation of the reaction: 2CH₃COOH + Mg(OH)₂ ---> Mg(CH₃COO)₂ + 2H₂O

Trial 1:

Moles of acetic acid = concentration * volume in litres

concentration of acetic acid = 0.88 M

volume of acid used = 7.8 mL = (7.8/1000) Litres = 0.0078 L

Moles acetic acid = 0.88 M * 0.0078 L

Moles acetic acid = 0.00686 moles

Moles of Mg(OH)₂:

From the equation of reaction, 2 moles of acetic acid reacts with 1 mole of Mg(OH)₂

Therefore, 0.00686 moles of acetic acid will react with 0.00686/2 moles of Mg(OH)₂ = 0.00343 moles of Mg(OH)₂

Moles of Mg(OH)₂ = 0.00343 moles

Neutralization capacity = moles of Mg(OH)₂/mass of milk of magnesia

Neutralization capacity = 0.00343 mole /2.5 g

Neutralization capacity = 0.00137 mol/g

Trial 2.

Moles of acetic acid = concentration * volume in litres

concentration of acetic acid = 0.88 M

volume of acid used = 8.2 mL = (8.2/1000) Litres = 0.0082 L

Moles acetic acid = 0.88 * 0.0082

Moles acetic acid = 0.00722 moles

Moles of Mg(OH)₂:

From the equation of reaction, 2 moles of acetic acid reacts with 1 mole of Mg(OH)₂

Therefore, 0.00722 moles of acetic acid will react with 0.00722/2 moles of Mg(OH)₂ = 0.00361 moles of Mg(OH)₂

Moles of Mg(OH)₂ = 0.00361 moles

Neutralization capacity = moles of Mg(OH)₂/mass of milk of magnesia

Neutralization capacity = 0.00361 mole /2.5 g

Neutralization capacity = 0.00144 mol/g

i believe the answer is a.

the answer is na (sodium)