This problem of the limiting reactant is:

Question

Given the reactant amounts specified in each chemical reaction, dtermine the limiting reactant in each case:
(I'm just going to take one problem out of the bunch}

HCl + NaOH yeilds NaCl + H2O
2.00 2.50
mol mol
(answers have to be in 3 significant figures)
2.50molNaOH x 1.00molHCl = 2.50 molof
1.00molNaOH HCl
So now we have to find the amount of moles of excess reactant that remains and that's where I get stuck

There is a short way and a long way. I like the long way because its easier to explain. After you are acquainted with the problems of this type you can switch to the short way.
1. You have the equation.
Step 2. Convert grams to mols but you have mol now.
Step 3a. Use the equation to convert mols of HCl to mols of either water or NaCl. Let's choose NaCl.
2.00 mols HCl x (1 mol NaCl/1 mol HCl) = 2.00 mols NaCl produced.

Step 3b. Do the same thing for NaOH.
2.50 mols NaOH x (1 mol NaCl/1 mol NaOH) = 2.50 mols NaCl produced.

Step 3c. Obviously, both answers can't be correct. The amount of product produced is always the smaller of the two numbers. In this case, we will produce 2.00 mols NaCl AND THIS MAKES HCl the limiting reagent AND IT MAKES NaCl the excess reagent.

Step 3d. Convert mols limiting reagent to mols of the excess reagent.
2.00 mols HCl x (1 mol NaOH/1 mol HCl) = 2.00 mols NaOH used.

Step 3e. How much NaOH remains?
2.50 mol initially - 2.00 mol used = 0.50 mol NaOH remains unreacted. And you are not allowed three significant figures in this case because neither HCl nor NaOH are given past the hundreths place.

The short way.
2.00 mol HCl reacts with 2.00 mols NaOH leaving 2.50 - 2.00 mol = 0.50 mol NaOH unreacted.

BTW, notice the correct spelling of yields. Or place an --> or ===> there.

Thanks sorry about the mistake in spelling