Question

What volume of \[\color{blue}{CO_2}\] is obtained in the combustion of two liters of butane? The volumes of both gases have been measured in identical conditions.

Answer 1

@mathslover

Answer 2

every molecule of butane has 4 carbon atoms (C4H10). Every molecule of CO2 has only 1. SO every time you burn 1 molecule of butane you'll make 4 molecules of CO2, assuming you have enough oxygen to do the job.

Avogadro's law states that equal volumes of gas contain equal molecules under identical conditions, so 2L of butane will produce four times as much CO2 under the same conditions, or 8L of CO2.

I can do this with a balanced reaction and unit conversions, if you like

Answer 3

ya actually i too want a balanced reaction.

Answer 4

a combustion reaction always has 4 pieces, and 3 of them are always the same. When a fuel is burned in O2, you always produce CO2 and H2O, so we start with:
\[C_4H_{10} + O_2 \rightarrow CO_2 + H_2O\]
balance C's first:
\[C_4H_{10} + O_2 \rightarrow 4CO_2 + H_2O\]
balance H's next:
\[C_4H_{10} + O_2 \rightarrow 4CO_2 + 5H_2O\]
now we run into a problem because counting the O's on the products side we get 9, which is odd. There's no way to put a whole number coefficient in front of the O2 and have it equal 9. We can do this:
\[C_4H_{10} + \frac{9}{2}O_2 \rightarrow 4CO_2 + 5H_2O\] but a lot of teachers don't like fractions in balanced reactions, so we usually double everything and get rid of the fraction, so we end up with:\[2C_4H_{10} + 9O_2 \rightarrow 8CO_2 + 10H_2O\]

Answer 5

@JFraser I think u r somewhere wrong I mean u typed wrong

Answer 6

at left side there must be 13 O atoms in ur second last para:) @j\

Answer 7

you are correct, there should be 13, not 9 but the odd # trick still works

Answer 8

k! thanx a lot.
I will do it now:)

Answer 9

I forgot to count the O's from the CO2

Answer 10

hm.......... i saw that.
I knew that u can't do wrong:)

Answer 11

@JFraser are u chem teacher?

Answer 12

i am

Answer 13

oh i see gr8.
thanx for ur gr8 help:)