How do you find the limiting reactant in a single replacement reaction?
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.
To find the limiting reactant in a single replacement reaction, you need to compare the amounts of each reactant and determine which one will be completely used up first.
Here are the steps to find the limiting reactant:
1. Write and balance the chemical equation for the reaction.
2. Convert the given amounts of each reactant to moles, if necessary.
3. Use stoichiometry to determine the number of moles of the product that can be formed from each reactant.
4. Compare the calculated amounts of the product. The reactant that produces the smaller amount of the product is the limiting reactant.
Let's illustrate these steps with an example:
Example: In the reaction between 2 moles of aluminum (Al) and 3 moles of copper(II) sulfate (CuSO4), determine the limiting reactant.
1. The balanced chemical equation is:
2 Al + 3 CuSO4 -> Al2(SO4)3 + 3 Cu
2. Convert the given amounts of each reactant to moles:
2 moles Al
3 moles CuSO4
3. Use stoichiometry to calculate the number of moles of the product that can be formed:
From 2 moles Al, using the stoichiometric ratio of 2:1, we find that 2 moles of Al can produce 1 mole of Al2(SO4)3.
From 3 moles CuSO4, using the stoichiometric ratio of 3:1, we find that 3 moles of CuSO4 can produce 1 mole of Al2(SO4)3.
4. Compare the calculated amounts of the product:
Since both reactants produce the same amount of product (1 mole of Al2(SO4)3), neither reactant is in excess. Therefore, neither reactant is the limiting reactant in this case.
However, if one reactant had produced a smaller amount of product, that reactant would be the limiting reactant.
To find the limiting reactant in a single replacement reaction, you need to compare the amount of available reactants to the stoichiometry of the balanced chemical equation. Here are the steps:
1. Write the balanced chemical equation for the single replacement reaction. Make sure that both sides of the equation have an equal number of atoms for each element.
2. Identify the reactant that will undergo replacement, which is usually an element. Also, identify the other reactant that will remain unchanged, which is usually a compound.
3. Determine the moles or mass of each reactant available. If given the mass, convert it to moles using the molar mass of the substance.
4. Calculate the theoretical yield of the product for both reactants. This can be done by using the stoichiometry of the balanced equation.
5. Compare the calculated theoretical yields. The reactant that produces the smaller yield is the limiting reactant.
A limiting reactant is the reactant that is completely consumed and determines the maximum amount of product that can be formed. It restricts the reaction from proceeding further.
It is important to note that the limiting reactant may not always be obvious based on the reactant quantities given. Therefore, a comparison using the stoichiometry of the balanced equation is essential to determine the limiting reactant accurately.