. Butane C4H10 undergoes combustion in the following reaction:

Question

2 C4H10 + 13 O2 8 CO2 + 10 H2O
In a reaction, 40.00 g of butane are reacted with 150.00 g of oxygen gas.
A) What mass of carbon dioxide is produced?

Answer 1

I have re-written the equation so it makes sense.
2C4H10 + 13O2 ==> 8CO2 + 10H2O
This is a limiting reagent (LR) problem. You know that when both (or more) amounts of the reactants are given. So I work these the long way.
mols each:
mols C4H10 = grams/molar mass = 40.00/58 = about 0.69
mols O2 = 150/32 = about 4.6

mols CO2 from C4H10 IF excess O2 = 0.69 x (8 mols CO2/2 mols C4H10) = about 2.76
mols CO2 formed from O2 IF excess C4H10 = 4.6 mols O2 x (8 mols CO2/13 mols O2) = 2.83

In LR problems, the smaller moles is always the winner because you can't get more than the smaller amount. So LR is C4H10 and you will be able to form 2.76 mols (but redo all the math since I estimated here and there).
Then grams CO2 formed = mols CO2 x molar mass CO2 = ?
Post your work if you run into trouble.

Answer 2

2 C4H10 + 13 O2 ----> 8 CO2 + 10 H2O ?
C4H10 = 4*12 + 10 = 58 g/mol so we have 40/58 mols = 0.69 mols
O2 = 2*16 = 32 g/mol so we have 150/32 mols = 4.69mols
your reaction requires mols O2/mols C4H10 = 13/2 = 6.5
we have mols O2 / mols C4H10 = 4.69/.69 = 6.8
so we have excess O2 and our experiment is ruled by the mols of C4H10
we get 8 mols of CO2 for every 2 mols of C4H10, 4 times
0.69 * 4 = 2.76 mols of CO2 which is 12 +32 = 44 g/mol
2.76 mols * (44 g/mol) = 121 g CO2

Answer 3

Oh well, Dr Bob already did it :)

Answer 4

Ahhh but we got the same answer and you showed how to do it the short way. I did it the long way.