What volume of oxygen at s.t.p is required for the complete combustion of methane according to the following equation. CH4 2O2= CO2 2H2O
You need to find the + sign on your computer. How much CH4 do you wish to combust? 1 mole CH4 requires 2 mols O2. 1 L CH4 requires 2 L O2.
To determine the volume of oxygen required for the complete combustion of methane (CH4) according to the given equation: CH4 + 2O2 → CO2 + 2H2O, we can use the stoichiometric coefficients.
From the balanced equation, we can see that 1 mole of methane (CH4) reacts with 2 moles of oxygen (O2) to produce 1 mole of carbon dioxide (CO2) and 2 moles of water (H2O).
At Standard Temperature and Pressure (STP), 1 mole of any gas occupies a volume of 22.4 liters.
Therefore, to determine the volume of oxygen required, we need to first calculate the number of moles of methane and then use the stoichiometric coefficients to find the corresponding number of moles of oxygen. Finally, we can convert the moles of oxygen into volume using the STP conversion factor.
Step 1: Calculate the number of moles of methane (CH4)
Given that the volume is not specified, we assume it to be 1 mole of methane.
Step 2: Use the stoichiometry to calculate the number of moles of oxygen (O2) required
From the balanced equation, we can see that 1 mole of methane reacts with 2 moles of oxygen. Therefore, if we have 1 mole of methane, we need 2 moles of oxygen.
Step 3: Convert the moles of oxygen to volume using STP conversion factor
Since 1 mole of any gas occupies a volume of 22.4 liters at STP, 2 moles of oxygen will occupy a volume of 2 * 22.4 = 44.8 liters.
Therefore, the volume of oxygen required for the complete combustion of methane, according to the given equation, is 44.8 liters.
To find the volume of oxygen required for the complete combustion of methane, we need to use the stoichiometry of the reaction equation.
The balanced equation for the combustion of methane is:
CH4 + 2O2 → CO2 + 2H2O
From this equation, we can see that one molecule of methane reacts with two molecules of oxygen to produce one molecule of carbon dioxide and two molecules of water.
Now we can use the molar ratios to calculate the volume of oxygen required.
1 mole of any gas at standard temperature and pressure (STP) occupies 22.4 liters.
The coefficient in front of oxygen (O2) in the balanced equation is 2, so we need 2 moles of oxygen for complete combustion of 1 mole of methane.
Therefore, 2 moles of O2 will require 2 x 22.4 liters = 44.8 liters of oxygen at STP for the complete combustion of CH4.