Amixture of methanol and propyl acetate contains 25.0 wt% methanol. (a) using a single dimensional equation, determine the g-moles of methanol in 200.0 kg of the mixture. (b) the flow rate of propyl acetate in the mixture is to be 100.0ib-moleh. what must the mixture flow rate be in lbm/h?

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:

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:

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

58750.08068 is the answer i think