. Butane C4H10 undergoes combustion in the following reaction:
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?
To find the mass of carbon dioxide produced, we need to calculate the stoichiometric coefficients in the balanced equation and use them to convert the given masses of butane and oxygen to the mass of carbon dioxide produced.
First, let's find the molar masses of butane (C4H10) and oxygen (O2):
Molar mass of butane (C4H10) = (4 x atomic mass of carbon) + (10 x atomic mass of hydrogen)
= (4 x 12.01 g/mol) + (10 x 1.01 g/mol)
= 58.12 g/mol
Molar mass of oxygen (O2) = 2 x atomic mass of oxygen
= 2 x 16.00 g/mol
= 32.00 g/mol
Next, let's calculate the number of moles of butane (C4H10) and oxygen (O2) given the masses provided:
Number of moles of butane (C4H10) = mass of butane / molar mass of butane
= 40.00 g / 58.12 g/mol
≈ 0.688 mol
Number of moles of oxygen (O2) = mass of oxygen / molar mass of oxygen
= 150.00 g / 32.00 g/mol
= 4.688 mol
Now, we can use the stoichiometric coefficients in the balanced equation to establish a proportion between the moles of butane and the moles of carbon dioxide:
From the balanced equation, we see that 2 moles of butane (C4H10) react to produce 8 moles of carbon dioxide (CO2). Therefore, the proportion can be set up as follows:
2 mol C4H10 / 8 mol CO2 = 0.688 mol C4H10 / x
Cross-multiplying and solving for x, we get:
x = (8 mol CO2 * 0.688 mol C4H10) / 2 mol C4H10
= 2.75 mol CO2
Finally, we can calculate the mass of carbon dioxide produced using the molar mass of carbon dioxide (CO2):
Mass of carbon dioxide = number of moles of carbon dioxide * molar mass of carbon dioxide
= 2.75 mol * (12.01 g/mol + 2 x 16.00 g/mol)
= 2.75 mol * 44.01 g/mol
≈ 121.03 g
Therefore, approximately 121.03 grams of carbon dioxide (CO2) would be produced when 40.00 grams of butane react with 150.00 grams of oxygen gas.
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.
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