posted by Chris Sung on .
Alumminum hydroxide reacts with sulfuric acid as follows: 2Al(OH)3+H2SO4-->Al2(SO4)+6H2O.
Which reagent is the limiting reactant when 0.500 mol Al(OH)3 and 0.500 mol H2SO4 are allowed to react?
How many moles of Al2(SO4)3 can form under these conditions?
How many moles of the excess reactant remain after the completion of the reaction?
Convert 0.5 mol Al(OH)3 to moles Al2(SO4)3.
Convert 0.5 mol H2SO4 to moles Al2(SO4)3.
It is likely the number for moles Al2(SO4)3 will not be the same so one of them must be wrong. The correct answer, in limiting reagent problems is ALWAYS the smaller value and the reagent producing that value is the limiting reagent.
After identifying the limiting reagent, convert moles of it to moles of the non-limiting reagent and subtract from the initial amount to identify the excess
Limiting reactant: H2SO4
0.167 moles of Al2(SO4)3 can be formed.
There are 0.167 moles of Al(OH)3 remaining.