How many grams of carbon dioxide ctsare produced when 78.0 grams of Methanol rea
To determine the number of grams of carbon dioxide produced when a given amount of methanol reacts, we need to use the stoichiometry of the balanced chemical equation for the reaction.
The balanced chemical equation for the reaction of methanol (CH3OH) with oxygen (O2) to produce carbon dioxide (CO2) and water (H2O) is as follows:
2 CH3OH + 3 O2 → 2 CO2 + 4 H2O
From the balanced equation, we can see that 2 moles of methanol react to produce 2 moles of carbon dioxide. Therefore, the stoichiometric ratio between methanol and carbon dioxide is 2:2 or 1:1.
Now, we can use this stoichiometric ratio to determine the amount of carbon dioxide produced:
1 mole of CH3OH = 32.04 grams (molecular weight of CH3OH)
1 mole of CO2 = 44.01 grams (molecular weight of CO2)
Given that you have 78.0 grams of methanol (CH3OH), we need to find the number of moles of CH3OH:
Moles of CH3OH = Mass of CH3OH / Molecular weight of CH3OH
Moles of CH3OH = 78.0 g / 32.04 g/mol ≈ 2.434 moles
Since the stoichiometric ratio between methanol and carbon dioxide is 1:1, the number of moles of CO2 produced will be the same as the number of moles of CH3OH:
Moles of CO2 = 2.434 moles
Finally, we can convert moles of CO2 to grams of CO2 using its molar mass:
Mass of CO2 = Moles of CO2 × Molecular weight of CO2
Mass of CO2 = 2.434 moles × 44.01 g/mol ≈ 107.15 grams
Therefore, approximately 107.15 grams of carbon dioxide are produced when 78.0 grams of methanol reacts.