The mixture flow rate in lbm/h = 117.65 lbm/h
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 :
[tex]\tt 15\%\times 200~kg=30~kg\\\\mol=\dfrac{mass}{MW}=\dfrac{30~kg}{32~kg/kmol}=0.9375~kmol[/tex]
mass of the methyl acetate(C₃H₆O₂,MW=74 kg/kmol,85% wt) in 200 kg :
[tex]\tt 85\%\times 200=170~kg\\\\mol=\dfrac{170}{74}=2.297~kmol[/tex]
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 :
[tex]\tt \dfrac{1~kg~mixture}{0.85~methyl~acetat}\times 100~lbm/h=117.65~lbm/h[/tex]
