Can anyone help me with another limiting reactant and percent yield problems I am having a hard time with this?

Let's look at 124. As with your other question, start by converting masses of each reactant into moles:

1142 g C6H5Cl / 112.56 g/mol = 10.145 mol C6H5Cl

485 g C2HOCl3 / 147.39 g/mol = 3.290 mol C2HOCl3

Another way to identify the limiting reactant is to use moles of one reactant to calculate how many moles of the other reactant would be required to react completely. Then, comparing that to what you have available will allow you to identify the limiting reactant.

10.145 mol C6H5Cl X (1 mol C2HOCl3 / 2 mol C6H5Cl) = 5.06 mol C2HOCl3

Comparing 5.06 mol chloral needed to react with all of the C6H5Cl to the moles you have availble tells you that chloral is the limiting reactant here, and C6H5Cl is present in excess.

Now, calculate the theoretical yield:

3.290 mol C2HOCl3 X (1 mol C14H9Cl5 / 1 mol C2HOCl3) = 3.29 mol C14H9Cl5

Mass C14H9Cl5 = 3.29 mol X 354.5 g/mol = 1166 g C14H9Cl5

% yield = (actual yield / theoretical yield) X 100

% yield = (200 g / 1166 g) X 100 = 17.1%

Seriously???

You've listed five problems to be solved, and posted essentially illegible photos, and not said a thing about what it actually is that you are having a "hard time" with. That's the key. What is it you don't understand? We fix that, and then you can do the work yourself.