Why didn't you show how you obtained that 1.676 L. I could have taken one look at it and told you where you went wrong.
How many mols is 1.4 grams butane. That's mols = g/molar mass = 1.40/58 = approx 0.024.
Convert mols C4H10 to mols CO2 produced. That's
0.024 mols C4H10 x (8 mols CO2/2 mol C4H10) = 0.024 x 4 = about 0.096 mols CO2. Plug that into
PV = nRT
1*V = 0.096 x 0.08206*(23+273)
Solve for V in liters.
Post your work if you get stuck.
Butane, C4H10, is a component of natural gas that is used as fuel for cigarette lighters. The balanced equation of the complete combustion of butane is:
2C4H10(g)+13O2(g)→8CO2(g)+10H2O(l)
At 1.00 atm and 23 ∘C, what is the volume of carbon dioxide formed by the combustion of 1.40 g of butane?
Express your answer with the appropriate units.
NOTE: I got 1.676L, but it says that it is a wrong answer :(
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