If 0.50L of gaseous CH4 is burned at STP, what volume of O2 is required for complete combustion?

Methane burns in oxygen to produce CO2 and H2O

CH4(g) + 2O2(g) --> 2H2O(l) + CO2(g)

And what volume of CO2 is produced?

To determine the volume of O2 required for complete combustion of 0.50L of CH4 at STP, we need to use the balanced chemical equation:

CH4(g) + 2O2(g) → 2H2O(l) + CO2(g)

From the equation, we can see that the ratio of CH4 to O2 is 1:2. This means that for every 1 mole of CH4 we need 2 moles of O2.

1 mole of gas at STP occupies 22.4L. Since we have 0.50L of CH4, we need to find the number of moles of CH4:

0.50L CH4 × (1 mol CH4/22.4L) = 0.0223 mol CH4

Now, using the ratio from the balanced equation, we can determine the number of moles of O2 needed:

0.0223 mol CH4 × (2 mol O2/1 mol CH4) = 0.0446 mol O2

Finally, to find the volume of O2 required, we'll need to convert the moles back to liters:

0.0446 mol O2 × (22.4L/mol) = 1.00L O2

Therefore, 1.00L of O2 is required for complete combustion of 0.50L of CH4 at STP.

Now let's determine the volume of CO2 produced. Based on the balanced equation, we can see that for every 1 mole of CH4 burned, we get 1 mole of CO2. So, the volume of CO2 produced will be equal to the volume of CH4 used, which is 0.50L.

Therefore, 0.50L of CO2 is produced during the complete combustion of 0.50L of CH4 at STP.

To determine the volume of O2 required for complete combustion of 0.50L of gaseous CH4, we first need to use the balanced equation to find the stoichiometric mole ratio between CH4 and O2.

From the balanced equation:
1 mole of CH4 requires 2 moles of O2.

Since we have the volume of CH4 and want to find the volume of O2, we can use the ideal gas law equation. At STP (standard temperature and pressure), 1 mole of any gas occupies 22.4L.

0.50L of CH4 = (0.50L * 1 mol CH4) / (22.4 L/mol CH4) = 0.0223 mol CH4

To find the volume of O2 required, we can use the stoichiometric mole ratio:

0.0223 mol CH4 × (2 mol O2 / 1 mol CH4) = 0.0446 mol O2

Finally, we can convert this back to volume using the ideal gas law equation:

(0.0446 mol O2) × (22.4 L/mol O2) = 0.99824 L

Therefore, approximately 0.998 L of O2 is required for complete combustion of 0.50 L of gaseous CH4 at STP.

To find the volume of CO2 produced, we can again use the stoichiometric mole ratio from the balanced equation:

1 mole of CH4 produces 1 mole of CO2

Since we have the number of moles of CH4 (0.0223), we can directly convert this to the volume of CO2 using the ideal gas law equation:

(0.0223 mol CO2) × (22.4 L/mol CO2) = 0.5 L

Therefore, approximately 0.5 L of CO2 is produced.

You've worked more difficult problems than this. What's the trouble here?