1) Alum is a compound used in a variety of applications including cosmetics, wat
ID: 543052 • Letter: 1
Question
1) Alum is a compound used in a variety of applications including cosmetics, water purification, and as a food additive. It can be synthesized from aluminum metal, sulfuric acid, water, and potassium hydroxide, as seen in the following equation:
2Al(g) + 2KOH(aq) + 4H2SO4(aq) + 10H2O =====> 2KAl(SO4) + 12(H2O) = 3H2
Using the data below, determine the theoretical and percent yield for this alum synthesis. Note that aluminum is the limiting reactant.
Bottle Mass - 10.0959 g
Bottle with Aluminum Pieces - 11.1050g
Final Product with Bottle Mass - 19.9793g
2) What is the best course of action when acid or base is spilled on the skin?
a) Rinse area w water, b) apply base to an acid burn or acid to base burn, c) remove as much acid or base as possible with a towel d) None of the above
3) Aluminum metal from beverage cans can be used to synthesize alum, which is used as a food additive and in water purification. The experimental procedure specifies that small pieces of aluminum be used in the synthesis. What effect would starting with larger pieces of aluminum have?
a) The reaction would not occur at all. Smaller pieces of aluminum have an exposed reactive inner core.
b) The reaction would proceed very slowly due to the low surface area, and would not finish during the designated lab time.
c) Since the aluminum cannot be bent for the reaction to proceed, the aluminum would not fit into the reaction vessel.
d) The reaction would proceed more quickly due to the aluminum's low surface area, causing an unstable and dangerous reaction.
e) The larger piece of aluminum would have too much surface area, which would cause the reaction to occur too vigorously to monitor effectively.
Explanation / Answer
1) The balanced chemical equation is given.
2 Al (s) + 2 KOH (aq) + 4 H2SO4 (aq) + 22 H2O (l) --------> 2 KAl(SO4)2.12H2O (s) + 3 H2 (g)
As per the stoichiometric equation,
2 mole Al = 2 mole KAl(SO4)2.12H2O
Mass of Al metal taken = (11.1050 – 10.0959) g = 1.0091 g.
Mass of alum obtained = (19.9793 – 11.1050) g = 8.8743 g.
Atomic mass of Al = 26.9815 g/mol; therefore, mole(s) of Al corresponding to 1.0091 g Al = (1.0091 g)/(26.9815 g/mol) = 0.037399 mole.
Mole(s) of KAl(SO4)2.12H2O expected = 0.037399 mole.
Molar mass of KAl(SO4)2.12H2O = [1*39.0983 + 1*26.9815 +2*(1*32.065 + 4*15.9994) + 12*(2*1.008 + 1*15.9994)] g/mol = 474.3898 g/mol.
Mass of KAl(SO4)2.12H2O expected = (0.037399 mole)*(474.3898 g/mol) = 17.7417 g. This is the expected theoretical yield.
Percent yield = (8.8743 g)/(17.7417 g)*100 = 50.0195% 50.02% (ans).
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