This is the way I think about "limiting reagents." A lot of people tell me I have a strange way of thinking, so I don't know if it'll work for you. I found this problem, let's use it as an example:
Mg3N2 + 6H2O
3Mg(OH)2 + 2NH3
Mg3N2 is 58.1g and the initial amount of H2O is 20.4g
Step 1:Analyze the "actual" ratio of the reactants.
For every Mg3N2 molecule there are six water molecules. So the mole ratio is 1:6=0.167 (the order of the ratio is important)
Step 2: Convert from grams-->moles
Mg3N2:
molar mass(Mg3N2)= 100.93 g/mol
moles= mass/molar mass= 58.1g/100.93 g/mol= 0.576 mol Mg3N2
Water:
molar mass(H2O) = 18.02 g H2O / mol H2O
moles= mass/molar mass= 20.4g/18.02 g/mol= 1.13 mol H2O
Step 3: Analyze the "present" ratio
0.576 mol Mg3N2/ 1.13 mol H2O =0.509
Step 4: Compare "Present" ratio to "actual" ratio
1. If present ratio> actual ratio, then the water is the limiting reagent.
2. If present ratio<actual ratio, then the water is in exess. So the Mg3N2 is the limiting reagent.
3. If present ratio=actual ratio, the there is no limiting reagent.
In this case, present(0.509)>actual ratio(0.167), so
water is the limiting reagent.
Hope that helps!!!