Question

New industrial plants for acetic acid react liquid methanol (Mm=32g/m) with carbon monoxide in the presen of a catalyst. In an experiment, 15g od methonol and 10g of carbon monoxide were placed in a reaction vessel. What is the theoretical yeild of acetic acid?

Answer 1

21.42 g

Step-by-step explanation:

To know this, you need to put the equation that is taking and it's the following:

CH3OH (Methanol) + CO ---------> CH3COOH (Acetic acid)

We already know the mm of methanol, and the mm for CO is:

AM of C = 12 g/mol

AM of O = 16 g/mol

AM of H = 1 g/mol

MM of CO = 12 + 16 = 28 g/m

For acetic acid: 12 + 3(1) + 12 + 16*2 + 1 = 60 g/mol

Now that we know the molar mass of each molecule, let's calculate the number of moles of methanol and CO, because it's easier to work with moles than the grams

moles of methanol = 15/32 = 0.469 moles

moles of CO = 10/28 = 0.357 moles

According to the equation 1 mole of methanol reacts completely with 1 mole of CO. However, in this case, we can see that methanol exceed the CO, so, the CO is the limitant reactant so:

moles reactants of methanol: 0.469 - 0.357 = 0.112 moles.

moles of acid produced: 0.357 moles

This is because we have a ratio of 1:1 between methanol and CO. As methanol exceed CO, the CO is consumed completely leaving a remanant of methanol behind. the whole moles of CO are used, and appears as product of acetic acid, therefore, the moles of acetic acid are the same of CO.

Now with the molar mass of acid, let's calculate the gram of acid, and yield:

g = 0.357 * 60 = 21.42 g of acetic acid.

This is the theorical yield of acetic acid.