A 500-g sample of Al2(SO4)3 is reacted with 450 g of Ca(OH)2. A total of 596 g of CaSO4 is produced. What is the limiting reagent in this reaction, and how many moles of excess reagent are unreacted

Question

aaronq · Answer

1. convert the masses (m) to moles (g) :

$$n=\frac{ m }{ M }$$

M = molar mass

2. write an equation and balance it (because you need to know the stoichiometric coefficients)

3. When you're finding the limiting reactant, simply divide each species moles by it's coefficient. The amount of moles you get are "normalized" meaning that they're in a 1:1 ratio with the other species..

For example:  $$n_{normalized}=\frac{ Al _{2}(SO_4)_3 }{ Coefficient_{Al _{2}(SO_4)_3} }$$

once you do that for both reactants, the one with less "normalized" moles is the limiting reactant (because it's limiting the yield of the reaction).