Excess = Garbage (aka: the hardest type of mole question)

¡Bienvenido al blog de PinchOfKCN! ¿Como estas? ¡Ahora, voy a escribir un pasaje para el tema "Excess and Limiting Reactants"! ¡Es muy muy interesante! ¿No?

*cough cough* Enough of speaking Spanish, now switching back to normal English. So how's everyone doing? I see many tired faces, have you guys been completely wiped out by the math midterm? Me too, and what's worse, I'll be having my Spanish midterm very soon, that's why I'm practicing some Spanish in the beginning of this blog. Hoho.
Now, back to our topic of the day:

Have you ever made pizza or birthday cakes? When you finished making, you probably have some left over ingredients, right? Those left over ingredients are exactly the same as the excess quantity after a chemical reaction occurs. On the other hand, a limiting reactant is the reactant which will be completely used up during the reaction.

Return of the Mole King
Now it is where the mathematical calculations take place. Consider the following question:

44.6g magnesium reacts with 56.3g oxygen gas produces magnesium oxide. Find:
a.) How many grams of MgO is produced
b.) The limited and excess reactant
c.) How much of the excess reactant is left over

1. First, we will have to write a balanced equation. If an equation is already given, remember to check if it is balanced or not before you move on to the next step!
2 Mg + 1O₂  →  2 MgO


2. Then, convert each reactant to MgO to find out which reactant is limited and which one is excess.
The route you may want to follow is: grams to mol, mol to mol (ratio), then mol to grams
Mg to MgO:  44.6g Mg  x   1 mol Mg  x  2 mol MgO  x  40.3g MgO  =  74.0g MgO
                         24.3g         2 mol Mg         1 mol MgO

3. Now, do the same thing with O2.
O2 to MgO: 56.3g O₂ x 1 mol O₂  x  2 mol MgO  x  40.3 MgO  = 142g MgO
                     32.0 g        1 mol O2      1 mol MgO

4. Next, compare the amount of product produced by each reactant, the one which produce MORE MgO will be excess reactant, and the one which produce LESS will be limited reactant.
In this case, oxygen is excess and magnesium is limited, where 74.0g MgO will be produced.

5. After that, we can calculate how much oxygen is used.
44.6g Mg  x  1 mol O₂  x  1 mol O2  x  32.0g       =  29.4g of O₂
             24.3g Mg      2 mol Mg    1 mol O2

6. Finally, subtract the grams of O2 used from how many O₂ you have at the beginning.
56.3g - 29.4g = 26.9g O₂ left in excess

These questions sure look complicated (or in fact they ARE complicated), but once you've mastered them, you'll score 100% on the next test.
Now, a quick update on the mole map:

Pretty scary, huh? It's alright, don't stress out too much, just practice more, and you'll be all prepared for next class' quiz.