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 :

Percent error can be calculated by the difference of the theoretical value and the measured value divided by the theoretical value multiplied by 100 percent.
% error = 27.26 - 27.2 / 27.26 x100
% error = 0.22%
A value close to zero would mean that the measured value is more or less near the actual value.