A mixture of methanol and methyl acetate contains 130 weight

A mixture of methanol and methyl acetate contains 13.0 weight percent methanol. Determine the number of gmols of methanol 190.0 kilograms of the mixture.

200x10³g(0.15) = 30,000 g CH4

mole: n = m/M = 0.85(200x10³g)/74.08 g/mol) = 2294.8 mol

C3H8O2

100lb-mol/h/2294.8 mol = 0.04357 lbm/h