CH4 + 2O2 -> CO2 + 2H2O

When 70 grams of oxygen react with methane, how many liters of CO2 gas are produced?

To find the number of liters of CO2 gas produced, we need to use the stoichiometry of the balanced chemical equation and convert the given mass of oxygen to the volume of CO2.

First, let's calculate the number of moles of oxygen (O2) using its molar mass:

Molar mass of O2 = 2 * Atomic mass of O
= 2 * 16.00 g/mol
= 32.00 g/mol

Number of moles of O2 = Mass of O2 / Molar mass of O2
= 70 g / 32.00 g/mol
≈ 2.19 mol

Now, using the stoichiometry of the balanced equation, we can determine the number of moles of CO2 produced. From the equation, we see that 1 mole of methane (CH4) reacts with 2 moles of oxygen (O2) to produce 1 mole of CO2:

Number of moles of CO2 = Number of moles of O2
= 2.19 mol

Finally, we can use the ideal gas law to convert the number of moles of CO2 to the volume of gas at standard temperature and pressure:

PV = nRT

Where:
P = Pressure (usually measured in atm)
V = Volume (usually measured in liters)
n = Number of moles
R = Ideal gas constant (0.0821 L·atm/(mol·K))
T = Temperature (usually measured in Kelvin)

Since this question does not provide the pressure and temperature conditions, we will assume standard temperature and pressure (STP), which is 0 degrees Celsius (273 K) and 1 atmosphere (1 atm). With these values, we can rearrange the equation to solve for volume:

V = nRT / P

Using STP conditions, we can substitute the known values:

V = (2.19 mol)(0.0821 L·atm/(mol·K))(273 K) / (1 atm)
≈ 48.21 L

Therefore, approximately 48.21 liters of CO2 gas are produced when 70 grams of oxygen react with methane.

Follow the steps in this worked example.

http://www.jiskha.com/science/chemistry/stoichiometry.html