What is the Limiting Reagent?

Question

If I have 1.0 M of HCL reacting with 1.0 M NaOH, and there is .02 L of HCL and .08 L of NaOH? How many mols of water form?

Answer 1

HCl + NaOH --> NaCl + H2O
Since the moles of HCl to moles of NaOH is 1:1, and both solutions have the same concentration, HCl is the limiting reagent as it has the smaller volume.
Moles of HCl given = 0.02 x 1.0 = 0.02 mole
Moles of NaOH given = 0.08 x 1.0 = 0.08 mole

Since HCl is the limiting reagent, it will determine number of moles of water formed.
Moles of water formed = moles of HCl used
(since the ratio of water to HCl is 1:1)
Thus moles of water produced is 0.02 moles

Hope this helps :)

Answer 2

To deal with limiting reactant problems, you first have to have a balanced equation for the reaction. So, for this reaction, the equation is:

HCl + NaOH --> NaCl + H2O

(Aside: you can use whatever form of the equation works best for you, such as a net ionic equation...)

From the equation, you can see that there is a 1:1 ratio of HCl to NaOH. In other words, 1 mole of HCl will react with 1 mol of NaOH to produce 1 mol of water and 1 mol of NaCl.

So now, you need to determine how many moles of each reactant you have. You've got 0.02 L of 1 M HCl, or 0.02 mol of HCl. You've also got 0.08 L of 1 M NaOH, or 0.08 L of NaOH. The question then becomes which reactant do you have the smallest amount of? Your 0.02 mol of HCl could let you produce 0.02 mol of water, but your 0.08 mol of NaOH could let you produce 0.08 mol of water. Your limiting reactant is the one which will allow you to produce the smallest amount of the product. In this case, HCl is your limiting reactant, and you will be able to produce 0.02 mol of water.