I don\'t understand how to calculate the moles part, I need step by step explana
ID: 591388 • Letter: I
Question
I don't understand how to calculate the moles part, I need step by step explanation. 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
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
Explanation / Answer
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
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