Pre-Lab Questions Show your work for all calculations. Remember to use the corre
ID: 714456 • Letter: P
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Pre-Lab Questions Show your work for all calculations. Remember to use the correct number of significant figures and units. You may need to use additional paper. 1. Both the KOH solution and the H SO4 solutions are labeled with the GHS pictogram shown What is the meaning of this pictogram and what safety precautions are required? (There is a link to an article explaining the GHS pictograms in the orientation to this lab course.) 2. a. Write out Equation 1. What is the mole-to-mole ratio between the aluminum metal that dissolves and the K[AI(OH)4a)product in this reaction? b. Write out Equations 2 and 3. What is the mole-to-mole ratio between the K[Al(OHa and the KAl(SO)g product in this reaction? c. What is the overall mole-to-mole ratio between the aluminum metal reactant and the alum product, KASO12 HO 3. Calculate the molar masses of the following materials: Al Al:03s) KAl(SO4) 12 H:0 4. a. If a student weighs out 0.482 grams of aluminum foil and dissolves it in KOH solution, how many moles of aluminum metal react? b. How many moles of KOH are necessary to react with the aluminum metal? c. Calculate the number of moles of KOH in 13.0 mL of 3.0 M KOH solution?Explanation / Answer
Synthesis of alum starting with Aluminium
1. Both KOH and H2SO4 are highly dangerous and corrosive in nature. Extreme care must be taken while handling these chemicals. If spilled on skin, or inhaled, wash with plenty of water and consult a physician immediately. These are also strong irritants. Wear proper safety clothing and safety goggles.
2. a. Reaction equation,
2Al(s) + 2KOH(aq) + 4H2SO4(aq) + 10H2O(l) ---> 2KAl(SO4)2.12H2O(s) [Alum] + 3H2(g)
b. 2K[Al(OH)4](aq) + 2H2SO4(aq) + 8H2O(l) ---> 2K[Al(SO4)2].12H2O(s)
c. mole to mole ratio of Al metal and alum product is 1 : 1.
3. Molar masses of,
Al = 27 g/mol
Al2O3(s) = 2 x 27 + 3 x 16 = 102 g/mol
KAl(SO4)2.12H2O = 39 + 27 + 2 x 32 + 8 x 16 + 24 x 1 + 12 x 16 = 474 g/mol
a. moles Al reacted = 0.482 g/27 g/mol = 0.018 mol
b. moles KOH needed to dissolve Al = 0.018 mol
c. molex KOH present = 3 M x 13 ml = 39 mmol
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