INSTRUCTIONS: Read this exam carefully! Section "A" with steps 1 - 10 is a practice problem worked out for you. Understand it and then use your actual lab data from the experiment about mass balance and burning magnesium. Apply the method to your ACTUAL lab data. Show ALL work. You will have 60 minutes. The exam is worth 100 points.
A. Reactions of Magnesium
2Mg + O2 ⇒ 2MgO
MgO + H2O ⇒ Mg(OH)2
Ksp = 1.5 x 10 -11
Mg(OH)2 + H2SO4 ⇒ MgSO4 + 2H2O
Mg(OH)2 + 2HCl ⇒ MgCl2 + 2H2O
1. Assume 0.010 g Mg. How many moles of Mg is that?
(moles = mass/At. Wt. = 0.010 g/24 amu = 4.16 x 10-4)
2. Mole ratio of Mg to MgO is 1:1. Therefore, 4.16 x 10-4mole of MgO is formed.
3. Formula weight of MgO = 40. Mass of MgO = (4.16 x 10-4)(40) = 0.0167 g.
4. Assume recovery of 0.008 g or 48%. (0.008 g/0.0167 = 0.48 = 48%)
5. When added to water, solution pH is 8.5. pH = - log [H+] so,
8.5 = - log [H+]
- 8.5 = log [H+]
antilog -8.5 = antilog (log [H+])
10-8.5 = [H+] = 10-9 x 10.5 = 3.16 x 10-9
Since Kw = [H+][OH-] = 1.0 x 10 -14
[OH-] = 1.0 x 10-14 / 3.16 x 10-9 = 0.316 x 10 (-14 – (-9)) = 3.16 x 10-1 x 10-5
[OH-] = 3.16 x 10-6 M = 3.16 x 10-6 moles per liter
6. Since formula is Mg(OH)2, and 3.16 x 10-6 moles per liter is the TOTAL [OH-] per liter, the concentration of Mg(OH)2 must be 3.16 x 10-6 moles/2 = 1.58 x 10-6 moles per liter. The mass of Mg(OH)2 is (1.58 x 10-6 )(24 + (2)(17)) = 9.16 x 10-5 g.
7. Acid (H2SO4 or HCl) was added. The concentration is 1 M (1 Molar or 1 mole per liter). It reacted with the dissolved Mg(OH)2 to form MgSO4 and water (or MgCl2 and water). As it reacted, more of the MgO was able to react with the water to form more Mg(OH)2 which then reacted with the acid, and so on util the MgO was all used up.
8. If the only Mg(OH)2 was the amount that originally dissolved, and it was in 50 ml water, not 1 liter, the amount of Mg(OH)2 that would have reacted would have been (50 ml / 1,000 ml per liter) (9.16 x 10-5 g) = 4.58 x 10-6 g.
9. However, since 0.008 g MgO was recovered and eventually reacted, the amount of MgSO4 possible for recovery is in a 1.0 : 1.0 ratio. (What would it be for MgCl2?) Thus, [(0.008g/(24+16) amu MgO per mole] x (24 + 32 + 64) amu MgSO4 = 0.024 g.
10. But you don’t know how much water there was. It was not measured. You do know the final weight. If the beaker weighed 45.57 grams, then the gross weight of MgSO4 with beaker would be 45.59 g.
B. Now, using the model above, solve the following:
1. You know how much Mg you started with. From your lab report, how much was that? _____________
2. You know how much MgO you recovered. From your lab report, how much was that? _____________
3. You know what the final weight was of either the MgCl2 or MgSO4 that you recovered at the end of the experiment. From your lab report, how much was that? _____________
4. You also know how many drops of acid you added. From your lab report, how much was that? _____________
5. How many ml of acid is that, based on your finding of the number of drops to make a ml? _____________
6. Now, calculate the following:
a. Moles Mg at start.
b. Weight of MgO theoretically possible.
c. Percentage of MgO actually recovered.
d. Weight of MgSO4 or MgCl2 theoretically possible.
e. Percentage of MgSO4 or MgCl2 actually recovered.
f. Moles of acid needed to react with the Mg(OH)2.
g. Moles of acid you actually used.
h. When you heated the MgO, some students reported that the pH increased to 9. What is the [OH-] concentration when the pH is 9?
C. In the next section of this exam, write and balance the following reactions using the algebraic method that I taught to you:
1. K4Fe(CN)6 + H2SO4 + H2O = K2SO4 + FeSO4 + (NH4)2SO4 + CO
2. KMnO4 + HCl = KCl + MnCl2 + H2O + Cl2
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