Stoichiometry

I'll post something later on when you have to actually find a limiting reactant, the problems I showed earlier didn't require it.

EDIT:
On my test these equations weren't balanced, but I assume since you're this far in Chem you should know how to do it.

 Question 1

H(3)PO(4) + 3KOH -> K(3)PO(4) + 3H(2)O

If you have 12.00 grams of each reactant, which is the limiting reactant?

 Answer

So, the best way to do this is just find out how much of each you need react.

(12.00gH(3)PO(4) / 1 ) | ( 1mol H(3)PO(4) / 98gH(3)PO(4) ) | (3molKoH / 1molH(3)PO(4) )
0.0713molH(3)PO(4) required to completely react with an excess of KOH

(12.00gKOH/ 1) | ( 1molKOH / 56.1gKOH ) | (1molH(3)PO(4) / 3molKOH )
3.66molKOH required to completely react with an excess of H(3)PO(4)

LR = KOH

 Explanation

Therefor, because you require more of KOH to react KOH will be your limiting reacting. Once you run out of KOH, you reaction stops, and because you run out of it faster than you do H(3)PO(4) that is what makes it the limiting reactant.

 Question 2

H(3)PO(4) + 3KOH -> K(3)PO(4) + 3H(2)O

How many grams of potassium phosphate can be made with the amounts given above(12.00g each reactant)?

 Answer

(12.00gKOH / 1) | ( 1molKOH / 56.1gKOH ) | (1molK(3)PO(4) / 3molKOH ) | (212.3gK(3)PO(4) / 1molK(3)PO(4))

15.144gK(3)PO(4)

 Explanation

So this is just another g->m->m->g kind of situation. However, the question doesn't specify which reactant to use because one of them is a limiting reactant. Therefore, you must use the limiting reactant to find out how much can be produced. This is because one the limiting reactant is completely used up, the reaction will stop and the excess doesn't matter.