How many litres of oxygen are required to react with 23 g of methane according
to the following equation?
CH4 + 2O2 Æ CO2 + 2H2O
how many moles of CH4 in 23g?
You will need twice that many moles of O2
each mole occupies 22.4L at STP.
It is good
To determine the amount of oxygen required to react with 23 g of methane, we need to use stoichiometry and the molar masses of the compounds involved.
Step 1: Calculate the molar mass of CH4 (methane)
The molar mass of carbon (C) is 12.01 g/mol, and the molar mass of hydrogen (H) is 1.01 g/mol. Methane has one carbon atom and four hydrogen atoms, so the molar mass is:
(1 * 12.01 g/mol) + (4 * 1.01 g/mol) = 16.05 g/mol
Step 2: Convert the mass of CH4 to moles
To convert grams to moles, divide the mass by the molar mass:
23 g / 16.05 g/mol = 1.43 mol of CH4
Step 3: Use the mole ratio from the balanced equation to find the moles of O2
From the balanced equation, we can see that for every 1 mole of CH4, we need 2 moles of O2.
So, 1.43 mol of CH4 * (2 mol of O2 / 1 mol of CH4) = 2.86 mol of O2
Step 4: Convert moles of O2 to litres
To convert moles to litres, we need to use the ideal gas law equation (PV = nRT), assuming standard temperature and pressure (STP: 0 °C or 273.15 K, and 1 atm pressure). At STP, 1 mole of any gas occupies 22.4 L.
So, 2.86 mol of O2 * 22.4 L/mol = 64.1 L of O2
Therefore, to react with 23 g of methane, you would need 64.1 litres of oxygen.
To find out how many liters of oxygen are required to react with 23 g of methane, we need to use stoichiometry.
Step 1: Calculate the number of moles of methane (CH4) using its molar mass.
The molar mass of methane (CH4) is calculated by adding the atomic masses of each element present in it:
C = 12.01 g/mol
H = 1.01 g/mol (4 hydrogen atoms in methane)
Total molar mass of CH4 = (12.01 g/mol) + (1.01 g/mol x 4) = 16.05 g/mol
To calculate the number of moles, divide the given mass (23 g) by the molar mass:
Number of moles of CH4 = 23 g / 16.05 g/mol ≈ 1.43 mol
Step 2: Use the stoichiometry of the balanced equation to determine the molar ratio of oxygen (O2) to methane (CH4).
From the balanced equation, we can see that the stoichiometric ratio between CH4 and O2 is 1:2. This means that for every 1 mole of CH4, we need 2 moles of O2.
Step 3: Calculate the number of moles of oxygen (O2) needed.
Number of moles of O2 = (Number of moles of CH4) x (Molar ratio of O2 to CH4)
Number of moles of O2 = 1.43 mol x 2 = 2.86 mol
Step 4: Convert the number of moles of oxygen (O2) to volume (liters) using the ideal gas law.
The ideal gas law is given by: PV = nRT
Since we are interested in the volume (V) and the equation can be rearranged to solve for V, we have:
V = (nRT) / P
Assuming standard temperature and pressure (STP) conditions:
T = 273 K (0 °C)
P = 1 atm
R is the ideal gas constant, which is 0.0821 L·atm/mol·K.
V = (2.86 mol x 0.0821 L·atm/mol·K x 273 K) / 1 atm
V ≈ 64.53 L
Therefore, approximately 64.53 liters of oxygen are required to react with 23 g of methane according to the given equation.