Please show how to solve the following questions:
2Al (s) + 2KOH (aq) + 4 H2SO4 (aq) + 22 H2O (l) --> 2KAl (SO4)2*12H2O (s) + 3H2 (g)
12.0 g of Al, 420 mL 2.0M KOH, 216 mL H2SO4 placed in 300 mL ice bath (water in excess)
1) What is the limiting reagent?
2) How much aluminum is necessary to produce 18g of alum?
3) By what factor must the mole amount of AL be reduced to produce 18 g alum?
4) What is the result of scaling the # moles of KOH used in the given procedure by this factor? What volume of 1.4M KOH contains this # moles?
5) What is the result of scaling the total # moles of H2SO4 used by this factor? What volume of 9.0M H2SO4 contains this # moles?
6) Scale the moles of sulfuric acid in a 54 mL portion of 10.0M H2SO4 by this factor. What volume of 9.0M H2SO4 contains this number of moles?
7) What is the result of scaling a 60 mL sample of distilled water by this factor?
8) What is the result of scaling down 300 mL 50/50 ethanol-water ice bath by this factor?
1) To determine the limiting reactant find the moles of each reactant. Then divide these moles byt the coefficient of each term (for Al you would divide by 2, for KOH divide by ...
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