Don’t need work to be shown. Just an answer
Ethanoic acid is produced according to the equation below. Calculate the mass of ethanoic acid that would be produced by the reaction of 600 g of carbon monoxide gas with sufficient CH3OH
So you don't need to know how to solve the problem You just want us to do the work and plop the answer up for you to copy it to a homework sheet or an exam and you can get credit without learning anything. Sorry. I don't do that.
Bob
first step: what is the equation?
then convert masses to moles and see how many moles of product are produced. Then convert back to mass. Easy peasy.
To calculate the mass of ethanoic acid produced, we need to determine the limiting reactant first.
Here's how you can do it step-by-step:
1. Write out the balanced equation for the reaction:
CO + 2 CH3OH -> CH3COOH + H2O
2. Calculate the molar mass of each substance:
CO: 12.01 g/mol + 16.00 g/mol = 28.01 g/mol
CH3OH: 12.01 g/mol + (4 × 1.01 g/mol) + 16.00 g/mol = 32.04 g/mol
CH3COOH: (2 × 12.01 g/mol) + (4 × 1.01 g/mol) + 16.00 g/mol = 60.05 g/mol
3. Calculate the number of moles for each reactant:
Moles of CO = mass / molar mass = 600 g / 28.01 g/mol
Moles of CH3OH = (600 g / 32.04 g/mol) / 2 (since the stoichiometric coefficient is 2)
4. Determine the limiting reactant:
Compare the number of moles of CO with CH3OH. The reactant with the smaller moles is the limiting reactant.
5. Use the stoichiometry of the balanced equation to calculate the moles of CH3COOH produced from the limiting reactant:
Since the stoichiometric coefficient of CO is 1, the number of moles of CH3COOH is equal to the moles of CO.
6. Calculate the mass of ethanoic acid produced using the calculated moles and molar mass:
Mass of CH3COOH = Moles of CH3COOH × molar mass
By following these steps, you can calculate the mass of ethanoic acid accurately.