Answer:
a) 1562.5 moles
b) 138.9 lbm/h
Explanation:
a) Methanol (CH₃OH) has a molar mass equal to 32 g/mol (12 of C + 4x1 of H + 16 of O). The mixture has 0.25 of methanol, so, transforming, in a single dimensional equation, the 200 kg (200000 g) of the mixture to moles of methanol:
[tex]200000 g mixture x \frac{0.25 methanol}{1 g mixture}x\frac{1 mol methanol}{32 g methanol}[/tex]
1562.5 moles of methanol
b) Propyl acetate (C₅H₁₀O₂) mas molar mass equal to 102 g/mol (5x12 of C + 10x1 of H + 2x16 of O). So its number of moles should be:
[tex]200000 g mixturex\frac{0.75 g propyl acetate}{1 g mixture}x\frac{1 mol propyl acetate}{102 g propyl acetate}[/tex]
1470.6 moles of propyl acetate.
So, the molar fraction (x) will be:
x = moles of propyl acetate/total moles
x = 1470.6/(1470.6+1562.5) = 0.72
So, the mixture flow rate (F) is:
0.72F = 100
F = 138.9 lbm/h
