Question

A) How many of the following are found in 1500 g of benzene (C₆H₆)?

kg C₆H₆

mol C₆H₆

lb-mole C₆H₆

mole (g-mole) C

mol H

g C

g H

molecules of C₆H₆

B) A mixture of methanol and methyl acetate contains 45.0 wt% methanol.

Determine the g-moles of methanol in 300.0 kg of the mixture.

The flow rate of methyl acetate in the mixture is to be 100.0 lb-mole/h. What is the mixture flowrate in lb_m/h?

Answer 1

Given 1500g C₆H₆ (Molecular weight = 78)

kg  C₆H₆ = 1500/1000 = 1.5kg

mol  C₆H₆ = 1500/78 = 19.23 gmol

lbmol  C₆H₆ = 19.23*0.0022 = 0.0424 lbmol

moles of C = 6*moles of  C₆H₆ = 6* 19.23 = 115.38 gmol

moles of H = 6*moles of  C₆H₆ = 6* 19.23 = 115.38 gmol

g of C = 115.38 gmol *12 = 1384.56g

g of H = 115.38 gmol *1 = 115.38g

molecules of C₆H₆ = 19.23 * 6.023*10²³ = 1.1583*10²⁵ molecules

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