Question

How many liters of water vapor can be produced if 8.9 liters of methane gas (CH4) are combusted, if all measurements are taken at the same temperature and pressure? Show all of the work used to solve this problem.

CH4 (g) + 2 O2 (g) yields CO2 (g) + 2 H2O (g)

Answer 1

@Abhisar

Answer 2

@Preetha

Answer 3

@sweetburger

Answer 4

@dan815

Answer 5

@Kbug

Answer 6

@Abhisar

Answer 7

This one will also be solved using PV=nRT

Answer 8

We can see from the equation that for each mole of CH4, 2 moles of oxygen is produced. And also pressure and temperature is same. We can write that


\(\sf P= \huge \frac{nRT}{8.9} = \frac{2nRT}{V}\),

Here V is the volume of oxygen formed. Solve the equation for v.

Answer 9

Any problem?

Answer 10

not yet @Abhisar you may continue

Answer 11

That's it. V will be the volume of oxygen. Solve the equation for v.

Answer 12

im not sure how to solve the equation @Abhisar

Answer 13

\(\sf \huge \frac{nRT}{8.9} = \frac{2nRT}{V}\)

Can you solve it for V now?

Answer 14

Cancel out the common terms on both sides.

Answer 15

4.5?

Answer 16

@dan815

Answer 17

@Abhisar

Answer 18

It's not 4.5 unless my math is really off, all you have to do is cancel out the like terms as someone has already shown above. And if you're having a problem with that just factor out nRT from each side and that should help you get your answer

Answer 19

Because "all measurements are taken at the same temperature and pressure" moles are proportionate to volume and therefore conversions between moles and L can be ignored - that is we don't need the ideal gas law.

We set up a ratio (as we normally would) except using liters instead of moles, and plug in the variables we know:

\(\sf \dfrac{L~of ~CH_4}{CH_4's ~coefficient}=\dfrac{L~of~H_2O}{H_2O's ~coefficient}\rightarrow \sf \dfrac{8.9~L}{1}=\dfrac{L~of~H_2O}{2}\)

We solve this algebraically and obtain: \(\sf L~of~H_2O=\dfrac{2*8.9~L}{1}\)