When 57.0 g calcium and 32.8 g nitrogen gas undergo a reaction that has a 95.0% yield, what mass of calcium ni?

Haa Boy! ...This is going to take a while.

First: Whenever you're given a stoiciometric problem, which is what this is, you should write a balanced equation for the reaction, so-

                    3Ca + N2 ----> Ca3N2

If you look at this, Chris, you'll see that for every 3 moles of calcium you need 1 mole of nitrogen. You really don't know how many moles of either element you have so you've got to calculate that first.  

                 57g x 1 mole/40g = 1.425 moles of calcium and-
                 32.8g x 1 mole/28g = 1.171 moles of nitrogen

question: how many moles of nitrogen do you need for 1 mole of calcium?
Answer: 1/3rd. Can you see that, Chris? If it's 3 to 1, then it's 1 to 1/3rd. Okay, if we use up all the calcium (1.425 moles) then we'll need 1/3rd as many moles of nitrogen or 1.425/3 = 0.475 moles of nitrogen. From our calculations we see that we have 1.171 moles of nitrogen which is way more than enough to use up all the moles of calcium, so we're going to have 1.171 moles - 0.475 = 0.696 moles of nitrogen left over.

Next, lets assume that we use up all the calcium and make as much Ca3N2 as we can. We have 1.425 moles of calcium but we need 3 calcium atoms for every 1 molecule of Ca3N2, so 1.425/3 = the moles of calcium nitride we can make. That's 1.425/3 =0.475 moles of Ca3N2. Calcium nitride weighs 148 g/mole and 0.475 moles weigh-
       
                   0.475moles x 148g/mole = 70.3 grams...

But we've got one more glitch to get through. 70.3 grams is the amount of calcium nitride we'd get if we got 100% yield. Since we only get 95% of that we only get 95% of 70.3 grams or 0.95 x 70.3g = 66.8 grams....and that's your answer. I sure hope you followed this, Chris, I did my best to walk you through the whole thing.