The mixture flow rate in lbm/h = 117.65 lbm/h
<h3>Further explanation</h3>
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 :
mass of the methyl acetate(C₃H₆O₂,MW=74 kg/kmol,85% wt) in 200 kg :
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 :
Answer:
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I think
C: 12.0107 g/mol ≅ 12.00 g/mol
H: 1.00784 g/mol ≅ 1.008 g/mol
O: 15.999 g/mol ≅ 16.00 g/mol
n(molar mass of CH2O)= 180
n.30=180
n=6
molecular formula: c6h12o6 glucose
