Question

Moles are what we are supposed to be solving based on the information from the table. No other information has been given.

Data Table 1: Analysis of Alka-Seltzer

	Trial 1	Trial 2
Mass of acetic acid	5 g	5 g
Density of acetic acid	1.00 g/mL
Concentration of acetic acid	0.88 M
Mass of Alka-Seltzer	3.21 g	3.21 g
Volume of NaOH, initial	5 mL	5 mL
Volume of NaOH, final	1.2 mL	1.8 mL
Volume of NaOH, total	6.2 mL	6.8 mL
Moles of acetic acid	0.005456 moles ?	moles
Moles of NaOH	moles	moles
Moles of NaHCO3	moles	moles
Moles NaHCO3 / g Alka-Seltzer	moles	moles

Data Table 2: Analysis of Tums

	Trial 1	Trial 2
Mass of acetic acid	5 g	5 g
Density of acetic acid	1.00 g/mL
Concentration of acetic acid	0.88 M
Mass of Tums	1.29 g	1.29 g
Volume of NaOH, initial	5 mL	5 mL
Volume of NaOH, final	2.0 mL	2.6 mL
Volume of NaOH, total	7 mL	7.6 mL
Moles of acetic acid	0.00616 moles	moles
Moles of NaOH	moles	moles
Moles of NaHCO3	moles	moles
Moles CaCO3 / g Tums	moles	moles

Data Table 3: Analysis of Milk of Magnesia

	Trial 1	Trial 2
Mass of milk of magnesia	2.5 g	2.5 g
Density of milk of magnesia	1.14 g/mL
Volume of acetic acid, initial	5 mL	5 mL
Volume of acetic acid, final	8.2 mL (added)	7.6 mL (added)
Volume of acetic acid, total	13.2 mL	12.6 mL
Concentration of acetic acid	0.88 M
Moles of acetic acid	moles	moles
Moles of Mg(OH)2	moles	moles
Moles Mg(OH)2 / g milk of magnesia	moles	moles

Answer 1

Calculations

Data table-1 : Analysis of Alka-seltzer tablet

NaHCO3 + CH3COOH --> CH3COONa + CO2 + H2O

Trial 1,

mass of acetic acid = 5 g

volume of acetic acid = 5 g/1 g/ml = 5 ml = 0.005 L

concentration of acetic acid = 0.88 M

moles of acetic acid = 0.88 M x 0.005 L = 0.0044 mmol

Volume NaOH added = 5 - 1.2 = 3.8 ml = 0.0038 L

missing molar concentration of NaOH in the above data table

For calculation sake we would assume molar concentration of NaOH = 0.5 M

[pl. note molarity of NaOH is assumed and not real value used in experiment. Feed correct value to get right answer]

moles NaOH used = 0.5 M x 0.0038 ml = 0.0019 mol

moles NaHCO3 in tablet = 0.0044 - 0.0019 = 0.0025 mol

moles NaHCO3/g alka-seltzer = 0.0025 mol/3.21 g = 0.0008 mol/g

Data table-2 : Analysis of tums

NaHCO3 + CH3COOH --> CH3COONa + CO2 + H2O

Trial 1,

mass of acetic acid = 5 g

volume of acetic acid = 5 g/1 g/ml = 5 ml = 0.005 L

concentration of acetic acid = 0.88 M

moles of acetic acid = 0.88 M x 0.005 L = 0.0044 mmol

Volume NaOH added = 5 - 2 = 3 ml = 0.003 L

missing molar concentration of NaOH in the above data table

For calculation sake we would assume molar concentration of NaOH = 0.5 M

[pl. note molarity of NaOH is assumed and not real value used in experiment. Feed correct value to get right answer]

moles NaOH used = 0.5 M x 0.003 ml = 0.0015 mol

moles NaHCO3 in tablet = 0.0044 - 0.0015 = 0.0029 mol

moles NaHCO3/g alka-seltzer = 0.0029 mol/1.29 g = 0.00225 mol/g

Data table-3 : Analysis of Milk of Magnesia

Mg(OH)2 + 2CH3COOH ---> (CH3COO)2Mg + 2H2O

Trial 1,

mass of acetic acid = 5 g

volume of acetic acid = 8.2 - 5 = 3.2 ml = 0.0032 L

concentration of acetic acid = 0.88 M

moles of acetic acid = 0.88 M x 0.0032 L = 0.002816 mmol

moles of Mg(OH)2 = 0.002816/2 = 0.001408 mol

moles Mg(OH)2/g milk magnesis = 0.001408 mol/2.5 g = 0.0005632 mol/g