Thursday
March 23, 2017

Post a New Question

Posted by on Friday, October 29, 2010 at 1:20am.

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?

  • Chemistry - , Friday, October 29, 2010 at 1:29am

    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

  • Chemistry - , Sunday, February 7, 2016 at 6:23pm

    Limiting reactant: H2SO4

    0.167 moles of Al2(SO4)3 can be formed.

    There are 0.167 moles of Al(OH)3 remaining.

Answer This Question

First Name:
School Subject:
Answer:

Related Questions

More Related Questions

Post a New Question