Answer:
4714.950 kilograms is the mass of a sample containing 45.0 kmol of methyl acetate.
Explanation:
Moles of methyl acetate =[tex]n_1[/tex]=45.0 kmol= 45000 mol
Mole percentage of methyl acetate = 60.0%
Total moles in the sample = n
[tex]60.0\%=\frac{45000 mol}{n}\times 100[/tex]
[tex]n=\frac{45000 mol}{60.0}\times 100=75000 mol[/tex]
Mole percentage of methyl alcohol = 20.0%
Moles of methyl alcohol = n_2
[tex]20.0\%=\frac{n_2}{75000 mol}\times 100[/tex]
[tex]n_2=15,000 mol[/tex]
Mass of methyl alcohol = [tex]n_2\times 32.04 g/mol[/tex]
=[tex]15000 mol\times 32.04 g/mol=480,600 g[/tex]
Mole percentage of acetic acid = 20.0%
Moles of acetic acid = n_3
[tex]20.0\%=\frac{n_3}{75000 mol}\times 100[/tex]
[tex]n_3=15,000 mol[/tex]
Mass of acetic acid= [tex]n_3\times 60.05 g/mol[/tex]
[tex]15000 mol\times 60.05 g/mol=900,750 g[/tex]
Mass of methyl methyl acetate= [tex]n_1\times 74.08 g/mol[/tex]
[tex]45000 mol\times 74.08 g/mol =3,333,600 g[/tex]
Mass of sample: 480,600 g + 3,333,600 g + 900,750 g = 4714950 g
4714950 g = 4714.950 kg
(1 kg = 1000 g)
