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

There is a process to follow when dealing with limiting reactant problems. Since you are usually given masses of reactants, first use the molar mass of each reactant to convert masses into moles. Then, using the coefficients in the balanced equation you can determine which reactant is the limiting reactant. Then, you can use that reactant to calculate the moles and masses of any of the products.

So looking at problem 122, you have three reactants. So first, take each one and calculate moles of each:

15.0 g C3H6 / 42.08 g/mol = 0.3124 moles C3H6

10.0 g O2 / 32.0 g/mol = 0.3125 mol O2

5 g NH3 / 17.0 g/mol = 0.294 mol NH3

Now, there are a couple of ways to identify the limiting reactant. One way is to use the moles of each reactant available to calculate the moles of one of the products that could be formed. I like that method when you have more than 2 reactants. So, using the coefficients in the balanced equation, relate moles of each reactant to moles of C3H3N that can be formed:

0.3124 mol C3H6 X (2 mol C3H3N / 2 mol C3H6) = 0.3124 mol C3H3N

0.3125 mol NH3 X (2 mol C3H3N / 2 mol NH3) = 0.3125 mol C3H3N

0.294 mol O2 X (2 mol C3H3N / 3 mol O2) = 0.196 mol C3H3N

As you can see, O2 gives the smallest yield of C3H3N, and so O2 is the limiting reactant for this problem.

Now, use that to calculate the mass of C3H3N that you can form.

0.196 mol C3H3N X (53.06 g/mol) = 10.4 g C3H3N

You should be able to do the other two problems following the same procedure as this one.

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.