Question

The flowrate of methyl acetate in a methanol-methyl acetate mixture containing 16.0 weight percent methanol is to be 135.0 lbmols/hour. What must the overall mixture flowrate be in lb,mass/hour?

Answer 1

Answer : The overall mixture flow rate is 11905.71 lb.mass/hour

Explanation :

As we are given that 16.0 weight percent methanol. That means, 16.0 g of methanol present in 100 g of mixture (methanol-methyl acetate).

Mass of methanol = 16.0 g

Mass of mixture = 100 g

Mass of methyl acetate = 100 - 16.0 = 84.0 g

Now converting flow rate of methyl acetate from lb.mols/hour to lb.mass/hour.

`\text{lb.mols/hour}=(\text{lb.mols/hour})* \text{Molar mass}=\text{lb.mass/hour}`

Molar mass of methyl acetate = 74.08 g/mols

So,
`135.0\text{ lb.mols/hour}=(135.0\text{ lb.mols/hour})* 74.08g/mols=10000.8\text{ lb.mass/hour}`

Now we have to calculate the overall mixture flow rate.

As, 84 g of methyl acetate flow rate = 10000.8 lb.mass/hour

So, 100 g of mixture flow rate =
`(100)/(84)* 10000.8\text{ lb.mass/hour}=11905.71\text{ lb.mass/hour}`

Therefore, the overall mixture flow rate is 11905.71 lb.mass/hour