Trial 1 Trial 2 Balanced Chemical Equation for the reaction between the active i
ID: 699377 • Letter: T
Question
Trial 1
Trial 2
Balanced Chemical Equation for the reaction between the active ingredient in Milk of Magnesia & acetic acid
The active ingredient in Milk of Magnesia is magnesium hydroxide, Mg(OH)2.
Mg(OH)2 + 2 HC2H3O2 ------> Mg(C2H3O2)2 + 2 H2O
Mass of Milk of Magnesia (g)
2.53
2.56
Density of Milk of Magnesia (g/mL)
1.14
Initial Volume of acetic acid soln. in the syringe (mL)
5.00
5.00
Final volume of acetic acid soln. in syringe (mL)
0.5
1.00
Volume of acetic acid soln. delivered (mL)
4.5
4.0
Molarity of acetic acid (mol/L)
0.88 mol acetic acid/1 L
Moles of acetic acid used in the titration (mol)
(volume of acetic acid delivered in L)*(molarity of acetic acid) = (4.5 mL)*(1 L/1000 mL)*(0.88 mole acetic acid/1 L) = 0.00396
(volume of acetic acid delivered in L)*(molarity of acetic acid) = (4.0 mL)*(1 L/1000 mL)*(0.88 mole acetic acid/1 L) = 0.00352
Moles of Mg(OH)2 (mol)
As per the stoichiometry of the reaction,
1 mole Mg(OH)2 = 2 mole HC2H3O2.
Therefore,
0.00396 mole acetic acid = (0.00396 mol HC2H3O2)*(1 mol Mg(OH)2/2 mol HC2H3O2) = 0.00198
As per the stoichiometry of the reaction,
1 mole Mg(OH)2 = 2 mole HC2H3O2.
Therefore,
0.00352 mole acetic acid = (0.00352 mol HC2H3O2)*(1 mol Mg(OH)2/2 mol HC2H3O2) = 0.00176
Moles of Mg(OH)2/g Milk of Magnesia
(0.00198)/(2.53) = 0.0007826
(0.00176)/(2.56) = 0.0006875
Average moles of Mg(OH)2/g Milk of Magnesia
½*(0.0007826 + 0.0006875) = 0.00073505 = 7.3505*10-4
Trial 1
Trial 2
Balanced Chemical Equation for the reaction between the active ingredient in Milk of Magnesia & acetic acid
The active ingredient in Milk of Magnesia is magnesium hydroxide, Mg(OH)2.
Mg(OH)2 + 2 HC2H3O2 ------> Mg(C2H3O2)2 + 2 H2O
Mass of Milk of Magnesia (g)
2.53
2.56
Density of Milk of Magnesia (g/mL)
1.14
Initial Volume of acetic acid soln. in the syringe (mL)
5.00
5.00
Final volume of acetic acid soln. in syringe (mL)
0.5
1.00
Volume of acetic acid soln. delivered (mL)
4.5
4.0
Molarity of acetic acid (mol/L)
0.88 mol acetic acid/1 L
Moles of acetic acid used in the titration (mol)
(volume of acetic acid delivered in L)*(molarity of acetic acid) = (4.5 mL)*(1 L/1000 mL)*(0.88 mole acetic acid/1 L) = 0.00396
(volume of acetic acid delivered in L)*(molarity of acetic acid) = (4.0 mL)*(1 L/1000 mL)*(0.88 mole acetic acid/1 L) = 0.00352
Moles of Mg(OH)2 (mol)
As per the stoichiometry of the reaction,
1 mole Mg(OH)2 = 2 mole HC2H3O2.
Therefore,
0.00396 mole acetic acid = (0.00396 mol HC2H3O2)*(1 mol Mg(OH)2/2 mol HC2H3O2) = 0.00198
As per the stoichiometry of the reaction,
1 mole Mg(OH)2 = 2 mole HC2H3O2.
Therefore,
0.00352 mole acetic acid = (0.00352 mol HC2H3O2)*(1 mol Mg(OH)2/2 mol HC2H3O2) = 0.00176
Moles of Mg(OH)2/g Milk of Magnesia
(0.00198)/(2.53) = 0.0007826
(0.00176)/(2.56) = 0.0006875
Average moles of Mg(OH)2/g Milk of Magnesia
½*(0.0007826 + 0.0006875) = 0.00073505 = 7.3505*10-4
Explanation / Answer
Data Table 3 (5 points) Note that the "Balanced Chemical Equations" row has been added to this data table. Data Table 3: Analysis of Milk of Magnesia Trial 1 Trial 2 Balanced Chemical Equation for the reaction between the active ingredient in Milk of Magnesia & acetic acid Mass of Milk of Magnesia 2.53 2.56 Density of Milk of Magnesia (o/mL) 1.14 g/mL Initial Volume of Acetic Acid Soln. in the Syringe (mL) 5.00 5.00 0.5 4.5 Final Volume of Acetic Acid Soln. in Syringe (mL) Volume of Acetic Acid Soln. Delivered (mL)* 4 Molarity of Acetic Acicd (mol/L) Moles of Acetic Acid used in the titration (mol)* 0.88 mol acetic acid 1L 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 Moles of Mg(OH)2 (mol)* Moles Mg(OH)2/g Milk of Magnesia* Average Moles Mg(OH)2/g Milk of Magnesia * denotes a calculated value Post Lab Activity 1. (1.5 pts) Considering the relative stoichiometry of the reactions of acetic acid with NaHCO3 and with
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.