[Magus]

Hello, i need some help with this one, i have tried dividing and applying conversion factors, but nothing seems to give a logical output 

Acetic acid is produced industrially by the direct combination of methanol with carbon monoxide


CH3OH + CO = CH3COOH

How many grams of methanol to react with excess carbon monoxide to prepare 5,000 g of acetic acid, if the expected return is 88%?

[sjb]

Hello, i need some help with this one, i have tried dividing and applying conversion factors, but nothing seems to give a logical output 

Acetic acid is produced industrially by the direct combination of methanol with carbon monoxide


CH3OH + CO = CH3COOH

How many grams of methanol to react with excess carbon monoxide to prepare 5,000 g of acetic acid, if the expected return is 88%?

If 5.000 g is 88%, how much is 100%? How many moles of acetic acid is this? How many moles of methanol do you need to make this much acid? What is the mass of this amount of methanol?

[discodermolide]

Hello, i need some help with this one, i have tried dividing and applying conversion factors, but nothing seems to give a logical output 

Acetic acid is produced industrially by the direct combination of methanol with carbon monoxide


CH3OH + CO = CH3COOH

How many grams of methanol to react with excess carbon monoxide to prepare 5,000 g of acetic acid, if the expected return is 88%?

Work it out in moles.
According to your equation
1 mole MeOH + 1 mole CO gives 1 mole Acetic acid.
Try it from there!

[awkko808]

If CH3 + CO -> CH3COOH, then CH3 and CH3COOH react and form in a 1:1 ratio. Thus:

mol(CH3) x .88 = mol(CH3COOH)

This expression says that 88% of the methanol reacted is used to produce acetic acid. To find the moles of CH3 needed, you need to work backwards since you are given the mass of product. So, given the mass of CH3COOH, find the moles of CH3COOH using its molar mass, plug it into the expression and then solve for moles of CH3.