How many liters of water vapor can be produced if 8.9 liters of methane has (CH4) are combusted, if all measurements are taken at the same temperature and pressure? Please answer & explain

You can do this the long way or the short way. The long way is

1. Write the equation and balance it.
2. Convert L CH4 to mols CH4 using PV = nRT.
3. Using the coefficients in the balanced equation, convert mols CH4 to mols H2O vapor.
4. Using PV = nRT convert mols H2O vapor to volume.

The short way:
1. Write and balance the equation.
CH4 + 2O2 ==> CO2 + 2H2O

2. Use L directly with the coefficients in the balanced equation. As long as the equation deals with gases and there is no P or T difference between start and finish, there is no need to go through the middle steps.
8.4 L CH4 x (2 mols H2O/1 mol CH4) = 8.4 x 2 = 16.8 L H2O vapor.

To answer this question, we need to understand the stoichiometry of the chemical reaction between methane (CH4) and oxygen (O2) to produce water vapor (H2O) and carbon dioxide (CO2).

The balanced equation for the combustion of methane is:
CH4 + 2O2 -> CO2 + 2H2O

From the balanced equation, we can see that 1 mole of methane reacts with 2 moles of oxygen to produce 2 moles of water vapor. We need to convert the given volume of methane into moles and then use the mole ratio to determine the volume of water vapor produced.

To convert volume to moles, we need to use the ideal gas law:
PV = nRT

Where:
P = pressure (assumed constant)
V = volume of gas
n = number of moles
R = ideal gas constant
T = temperature

Since the measurements are taken at the same temperature and pressure, we can assume that these variables cancel out, leaving us with the equation:
V = n

The molar volume of any ideal gas at standard temperature and pressure (STP) is 22.4 liters/mol. Therefore, we can use this conversion factor to convert the given volume of methane into moles.

8.9 liters CH4 * (1 mol CH4 / 22.4 liters CH4) = 0.396 mol CH4

Now, we can use the mole ratio from the balanced equation to find the volume of water vapor produced.

0.396 mol CH4 * (2 mol H2O / 1 mol CH4) * (22.4 liters H2O / 1 mol H2O) = 17.73 liters H2O vapor

Therefore, approximately 17.73 liters of water vapor can be produced when 8.9 liters of methane is combusted under the given conditions.

Note: In this explanation, we assumed that methane and water vapor behave ideally as gases, which means they follow the ideal gas law. Additionally, the given volume of methane is assumed to be at the same temperature and pressure as the resulting water vapor.