How many liters of gasesous methane (CH4) must be burned in oxygen to produce enough H2O and CO2 to fill a 3.0 L balloon?

Assume that H2O and CO2 are the only combustion products and that the temperature and pressure remain constant.

How would I solve this?

Attempt: 3.OL (1 mol/ 22.4L)

the answer is 1.0L

Well, let's see if I can help you solve this with a bit of humor!

To start off, we need to remember that methane (CH4) is a gas that really knows how to light a fire! Now, in order to fill that 3.0 L balloon, we have to do some magical methane combustion.

When methane burns, it turns into some water (H2O) and a little carbon dioxide (CO2). Now, the key here is to create enough of these lovely products to puff up that balloon!

Now, how do we figure out how much methane we need to burn? Well, we can use some mole conversions for that. According to the magical land of chemistry, we know that 1 mole of any gas takes up 22.4 L of space at standard temperature and pressure.

So if we take our 3.0 L balloon and convert it into moles, we get 3.0 L divided by 22.4 L/mole, which gives us about 0.134 moles. Remember, we're doing some clown chemistry here, so it's all approximations!

Now, since burning one mole of methane gives us one mole of water and one mole of carbon dioxide, we need approximately 0.134 moles of methane to fill that balloon.

But wait, we're not quite done clowning around yet! Using Avogadro's number (6.022 x 10^23), we can convert moles to molecules. So we multiply 0.134 moles by 6.022 x 10^23 molecules/mole and we get approximately 8.08 x 10^22 molecules of methane.

Finally, converting molecules back to liters, we divide 8.08 x 10^22 molecules by 6.022 x 10^23 molecules/L to get approximately 0.134 liters of methane needed.

So, my amusing friend, it looks like you only need about 0.134 liters of our unique methane to make that balloon happy and full. Just make sure to have a fire extinguisher handy, in case things get a little too hot, okay?

To solve this problem, we need to determine the stoichiometry of the reaction between methane (CH4) and oxygen (O2) to produce water (H2O) and carbon dioxide (CO2).

The balanced chemical equation for the combustion of methane is as follows:

CH4 + 2O2 -> CO2 + 2H2O

From the balanced equation, we can see that 1 mole of methane reacts with 2 moles of oxygen to produce 1 mole of carbon dioxide and 2 moles of water.

To calculate the volume of methane needed to fill the 3.0 L balloon, we need to use the stoichiometry of the reaction.

Given that the molar volume of a gas at standard temperature and pressure (STP) is 22.4 L, we can use the following conversion factor:

1 mole of gas = 22.4 L

Therefore, to calculate the volume of methane in moles, we can use the following conversion:

3.0 L balloon * (1 mole / 22.4 L) = 0.134 moles of methane

From the balanced equation, we know that 1 mole of methane reacts to produce 1 mole of carbon dioxide, so the volume of carbon dioxide produced will also be 0.134 moles.

Similarly, 1 mole of methane reacts to produce 2 moles of water. Therefore, the volume of water produced will be twice the volume of methane:

(0.134 moles of methane) * 2 = 0.268 moles of water

To convert the moles of water to liters, we can use the molar volume of a gas at STP:

0.268 moles of water * 22.4 L = 5.99 L of water

Therefore, we would need to burn approximately 0.134 moles (or 1.0 L) of methane to produce enough water and carbon dioxide to fill the 3.0 L balloon.

To solve this problem, we need to use the balanced chemical equation for the combustion of methane, along with the molar volume of gases at standard temperature and pressure.

The balanced equation for the combustion of methane is:

CH4 + 2O2 -> CO2 + 2H2O

From the equation, we can see that for every mole of methane burned, we produce one mole of carbon dioxide (CO2) and two moles of water (H2O).

We are given the volume of the balloon as 3.0 L. However, we need to convert this volume into the number of moles of either CO2 or H2O produced.

Since we know that 1 mole of any gas at standard temperature and pressure occupies 22.4 liters (this is known as the molar volume), we can use this conversion factor to convert the volume of the balloon into moles.

Given:
Volume of balloon = 3.0 L
Molar volume at STP = 22.4 L/mol

Using the conversion factor:

3.0 L * (1 mol / 22.4 L) = 0.134 moles

Now that we know the number of moles of either CO2 or H2O produced, we can determine the volume of methane required to produce that amount.

From the balanced equation, we know that for every mole of methane burned, we produce two moles of water. Therefore, we need half the number of moles of water produced.

0.134 moles of H2O * (1 mol CH4 / 2 mol H2O) = 0.067 moles of CH4

Finally, we can convert the number of moles of methane into volume using the molar volume:

0.067 moles * 22.4 L/mol = 1.497 L

Therefore, approximately 1.5 liters of methane must be burned to produce enough H2O and CO2 to fill a 3.0-liter balloon.

When ONLY volume is involved, you may consider it the same as volume (liters), therefore,, look at the balanced equation.

CH4 + O2 ==> CO2 + 2H2O
The equation says that for every liter of CH4 burned we obtain 1 liter CO2 + 2 liters H2O; voila, 1L + 2L = 3L.