Question

A mixture of methanol and methyl acetate contains 15.0 wt% methanol.

The flow rate of the methyl acetate in the mixture is to be 100 lbm/h. What must be the mixture flow rate in lbm/h?

Answer 1

The mixture flow rate in lbm/h = 117.65 lbm/h

Further explanation

Given

15.0 wt% methanol

The flow rate of the methyl acetate :100 lbm/h

Required

the mixture flow rate in lbm/h

Solution

mass of methanol(CH₃OH, Mw= 32 kg/kmol) in mixture :

`\tt 15\%* 200~kg=30~kg\\\\mol=(mass)/(MW)=(30~kg)/(32~kg/kmol)=0.9375~kmol`

mass of the methyl acetate(C₃H₆O₂,MW=74 kg/kmol,85% wt) in 200 kg :

`\tt 85\%* 200=170~kg\\\\mol=(170)/(74)=2.297~kmol`

Flow rate of the methyl acetate in the mixture is to be 100 lbm/h.

1 kg mixture = 0.85 .methyl acetate

So flow rate for mixture :

`\tt (1~kg~mixture)/(0.85~methyl~acetat)* 100~lbm/h=117.65~lbm/h`