__C4H10 (g) + ___O2 (g) --> ___CO2 (g) + ___H2O (g)
So our first task is to balance this equation. Remember when we are balancing chemical equations the goal is to have the same number of each type of atom on each side. We can accomplish this only by changing the coefficients in front of each reactant or product, not the subscripts in the formulas themselves.
It's a good idea to start with either carbon or hydrogen. Let's leave oxygen for last because it appears in two products rather than just one. I'll start with carbon.
We've got 4 carbons on the left, to get 4 on the right we can put 4 in front of carbon dioxide.
__C4H10 (g) + ___O2 (g) --> _4_CO2 (g) + ___H2O (g)
Now hydrogen. We've got 10 on the left, we need 10 on the right, since each water molecule has 2, we would need a 5 in front of water.
__C4H10 (g) + ___O2 (g) --> _4_CO2 (g) + _5_H2O (g)
Finally oxygen. How many are on the right side of our equation? 4*2 + 5 = 13. Is it possible to achieve 13 oxygen atoms on the left? Ah, looks like we can't. We'll always have an even number of O atoms on the left no matter what coefficient we put in front of O2. You'll probably run into this problem while balancing equations many times. The solution is to multiply everything by 2, so that we have an even number again, without unbalancing carbon or hydrogen.
_2_C4H10 (g) + ___O2 (g) --> _8_CO2 (g) + _10_H2O (g)
Now how many oxygen atoms on the right? 2*8 + 10 = 26. How do I get 26 O atoms on the left? 13 in front of O2.
_2_C4H10 (g) + _13__O2 (g) --> _8_CO2 (g) + _10_H2O (g)
Now the reaction is balanced! We know the stoichiometry of the reaction now, or the ratio at which our reactants react and our products form. This 2:13:8:10 ratio applies to moles and to numbers of atoms, but not to masses!
So we have 4.50 L of butane (at STP) and 1.20 moles of oxygen. We need to know which reactant is going to run out first, because that will determine how much CO2 we can produce. We call this the limiting reactant. So let's see how many moles of CO2 I could make from each.
To figure out how many moles of butane we start with, we have two choices. One, we can remember that at standard temperature and pressure, an ideal gas will occupy 22.4 L per mole. If you're having trouble remembering that during an exam, you can also just use the ideal gas law, keeping in mind that standard temperature and pressure are 1 atm and 273.15 K, and the ideal gas constant is 0.08206 L atm / mol K.
So we can make less CO2 from the O2 than from the C4H10. Looks like the O2 is the limitting reagent, and the maximum number of moles of CO2 we can make is 0.738 mol.
Now for the percent yield, we'll need to know the theoretical yield of CO2 in grams. This is the maximum amount of CO2 we can produced based on the stoichiometry of the reaction and the amount of limitting reactant we have available. So we need to take our 0.738 mol of CO2 and convert it to grams. We need the molar mass of CO2. Hope you have a periodic table handy.
Percent yield is the fraction of the theoretical yield that we actually produced in the experiment.
