Question

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

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 :

$\tt 15\%\times 200~kg=30~kg\\\\mol=\dfrac{mass}{MW}=\dfrac{30~kg}{32~kg/kmol}=0.9375~kmol$

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

$\tt 85\%\times 200=170~kg\\\\mol=\dfrac{170}{74}=2.297~kmol$

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 :

$\tt \dfrac{1~kg~mixture}{0.85~methyl~acetat}\times 100~lbm/h=117.65~lbm/h$