Aluminum hydroxide reacts with sulfuric acid. Whichreagent is the limiting react

Question

Aluminum hydroxide reacts with sulfuric acid. Whichreagent is the limiting reactant when 0.700 mol of aluminumhydroxide and 0.700 mol sulfuric acid are allowed toreact?

How many moles of Al2(SO4)3 can form under theseconditions?

How many moles of the excess reactant remain after thecompletion of the reaction?

I need help with this problem. I know that the balancedequation is 2Al(OH)3+3H2SO4--->Al2(SO4)3 + 6H20

I keep getting 22.8 and the answer is incorrect. Cananyone help?

Explanation / Answer

               2Al(OH)3+3H2SO4--->Al2(SO4)3 + 6H20 2 moles of aluminum hydroxide react with 3 moles of sulfuricacid For 0.7 moles of aluminum hydroxide we require 1.05 moles ofsulfuric acid. But we have only 0.7 moles of sulfuric acid. So sulfuric acid is the limiting reagent . one mole of aluminum sulfate is formed from 3 moles of sulfuricacid. So moles of aluminum sulfate formed                                    = 0.7moles sulfuric acid * ( 1 mole Al2(SO4)3 / 3 moles acid)                                    = 0.233 moles . Since sulfuric acid is the limiting reactant, 0.7 moles sulfuricacid react with 0.7 * 2 /3 =0.47 moles of aluminum hydroxide. So excess moles of reactant = 0.7 - 0.47 = 0.329 moles

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