How do you find the limiting reactant in a single replacement reaction?

Answer 1

The procedure is the same no matter what kind of reaction and there is a long way as well as a short way. I'll show you both. For example, let's take a single replacement reaction such as
Cu(s) + 2AgNO3(aq) ==> 2Ag(s) + Cu(NO3)2(aq)
Let's make the numbers easy. Suppose we have 32 g Cu and 17 g AgNO3. To make the numbers easy I'll use 64 for the atomic mass of Cu and 170 as the molar mass of AgNO3.
Long way:
mols Cu = 32/64 = 0.5 mole to start
mols AgNO3 = 17/170 = 0.1 mol to start. Now go through two separate stoichiometry calculations with Cu to see how many moles Ag can be produced. That will be 0.5 mol Cu x (2 mol Ag/1 mol Cu) = 01.0 mol Ag produced from the 0.5 mol Cu. Do the same with AgNO3. You can produce with AgNO3 0.1 mols AgNO3 x (2 mol Ag/2 mol AgNO3) = 0.1 mol. In limiting reagent problems the small number ALWAYS wins; i.e., you can always proeduce the SMALLER AMOUNT. Common sense tells you you can't produce more than the smaller amount in the process. The short way:
mols Cu = 64/32 = 0.5 mol to start
mols AgNO3 = 17/170 = 0.1 mol to start.
Ask yourself. Starting with 0.5 mol Cu, how many moles AgNO3 will I need to react completely with that 0.5 mol Cu. The answer is
0.5 mol Cu x (2 mols AgNO3/1 mol Cu) = 1.0 mol AgNO3 needed to react with all of the Cu. Do you have that much AgNO3. No, you have ONLY 0.1 mol AgNO3; therefore, AgNO3 is the limiting reagent. Let me point out that you don't need to use Cu to start. Let's start with AgNO3. We have 0.1 mol AgNO3. We ask how much Cu will we need to react completely with our 0.1 mol AgNO3. The answer is 0.1 mol AgNO3 x (1 mol Cu/2 mols AgNO3) = 0.05 mols Cu. Do we have that much Cu. Yes we do; therefore, we have enough Cu so AgNO3 is the limiting reagent. Limiting reagent problems are not hard; you just remember they are two stoichometric problems rolled into one problem.