what exactly is bothering you about those problems? identifying the limiting reagent? If so, let's say they give you the equation:
2A + 3B -------> 5C + 2D
You're given 4 grams of A and 5 grams of B, who is the limiting reagent, and why?
You convert A and B into moles (grams/MW=moles)
Let's say you end up getting 1 mole for A and 2 moles for B. We can do it like this:
put the value you got for A under A in the equation. Cross multiply to see how much B that would give you (1x3/2=3/2). You would need 3/2 moles of B for that one mole of A. Our calculation gave us 2 moles for B, so we have more than enough of B, right? Therefore, A is the limiting reagent.
You can also do it using B, and seeing whether you have enough A. You'll get the same answer.
You can also divide the # of moles of A you calculated by the coeffficient of A in the equation. Do the same with B. Compare the two values, and the smaller one is the limiting reagent.
Hopefully I didn't make this even more confusing. Best of luck