Stoichiometry: Calculations with Chemical Formulas and Equations

 

70) 2Al(OH)3(s) + 3H2SO4(aq) Al2(SO4)3(aq) + 6H2O(l)

       n = 0.450 mol Al(OH)3          n = 0.550 mol H2SO4

     

       n = 0.450 mol Al(OH)3 * 1 mol Al2(SO4)3/2 mol Al(OH)3 = 0.225 mol Al2(SO4)3

 

       n = 0.550 mol H2SO4 * 1 mol Al2(SO4)3/3 mol H2SO4 = 0.183 mol Al2(SO4)3  

       H2SO4 produces the least amount of Al2(SO4)3, therefore it is the limiting reactant.

 

       n = 0.183 mol Al2(SO4)3 * 2 mol Al(OH)3/1 mol Al2(SO4)3 = 0.367 mol Al(OH)3

       n = 0.450 mol Al(OH)3 - 0.367 mol Al(OH)3 = 0.083 mol Al(OH)3 remains.