Methane is oxidized to water and carbon dioxide in a process, in which 10 % of the methane is converted to CO and 20 % is converted to formaldehyde. What is the yield of the process? How much methane do you need to produce a kg of formaldehyde? How much oxygen is needed in the process?
To calculate the yield of the process, we need to know the percentage of methane that is converted to the desired products (water and carbon dioxide). In this case, 10% of the methane is converted to carbon dioxide (CO2) and 20% is converted to formaldehyde (HCHO).
To calculate the yield, we add up the conversion percentages for both products:
Yield = CO2 conversion + HCHO conversion
Yield = 10% + 20%
Yield = 30%
So, the yield of the process is 30%.
To find out how much methane is needed to produce a kilogram of formaldehyde, we need to know the molar mass of methane (CH4) and formaldehyde (HCHO).
The molar mass of methane (CH4) is:
C (12.01 g/mol) + 4 H (1.01 g/mol) = 16.05 g/mol
The molar mass of formaldehyde (HCHO) is:
H (1.01 g/mol) + C (12.01 g/mol) + O (16.00 g/mol) = 29.02 g/mol
Now, we can set up a proportion to find out the amount of methane needed to produce a kilogram (1000 g) of formaldehyde:
(16.05 g CH4) / (1000 g HCHO) = x / (29.02 g HCHO)
Cross-multiplying and solving for x:
x = (16.05 g CH4 * 1000 g HCHO) / 29.02 g HCHO
x ≈ 554.6 g CH4
So, approximately 554.6 grams (or 0.5546 kg) of methane is needed to produce a kilogram of formaldehyde.
To calculate the amount of oxygen needed in the process, we need to consider the stoichiometry of the reaction. The balanced equation for the oxidation of methane to water and carbon dioxide is:
CH4 + 2O2 → CO2 + 2H2O
From the balanced equation, we can see that for every one mole of methane, we need two moles of oxygen.
To calculate the amount of oxygen needed, we need to find out the number of moles of methane required. Since we already know the molar mass of methane (16.05 g/mol), we can divide the mass of methane needed (0.5546 kg) by the molar mass:
Moles of CH4 = (mass of CH4) / (molar mass of CH4)
Moles of CH4 = (0.5546 kg * 1000 g/kg) / 16.05 g/mol
Moles of CH4 ≈ 34.55 mol
Since the stoichiometric ratio between methane and oxygen is 1:2, we need twice as many moles of oxygen:
Moles of O2 = 2 * (moles of CH4)
Moles of O2 = 2 * 34.55 mol
Moles of O2 ≈ 69.10 mol
Therefore, approximately 69.10 moles of oxygen are needed in the process.