1. Sulfuric acid and aluminum hydroxide react as follows: (balance the equation)
ID: 828532 • Letter: 1
Question
1. Sulfuric acid and aluminum hydroxide react as follows: (balance the equation)
__H2SO4 + __Al(OH)3 --> __H2O + __Al2(SO4)3
a. If 4.57 moles of sulfuric acid and 6.57 moles of aluminum (III) hydroxide react to completion ( until one of them is used up), show which reagent is the limiting reagent.
b. What is the greatest amount of moles of Al2(SO4)3 that can be formed under these conditions.
2. Aluminum burns in oxygen to form aluminum oxide. Write the correct balanced chemical equation for this reaction.
a. If 67 g of aluminum and 58 g of oxygen react to completion (until one of them is used up), show which is the limiting reagent.
b. What is the greatest amount of moles of aluminum oxide that can be formed under these conditions?
*please explain how you do it, and thank you so much!!
Explanation / Answer
(1) (a) 3 H2SO4 + 2 Al(OH)3 => 3 H2O + Al2(SO4)3
Theoretical moles of H2SO4 : Al(OH)3 = 3 : 2
Actual moles of H2SO4 : Al(OH)3 = 4.57 : 6.57 = 1 : 1.44 = 3 : 4.31
Since Al(OH)3 is in excess => H2SO4 is the limiting reagent
(b) Moles of Al2(SO4)3 = 1/3 x moles of H2SO4
= 1/3 x 4.57 = 1.52 mol
(2) (a) 4 Al + 3 O2 => 2 Al2O3
Theoretical moles of Al : O2 = 4 : 3
Moles of Al = mass/molar mass of Al = 67/26.98 = 2.483 mol
Moles of O2 = mass/molar mass of O2 = 58/32.00 = 1.8125 mol
Actual moles of Al : O2 = 2.483 : 1.8125 = 1.37 : 1 = 4.11 : 3
Since Al is in excess => O2 is the limiting reagent
(b) Moles of Al2O3 = 2/3 x moles of O2
= 2/3 x 1.8125 = 1.21 mol
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