Question

A mixture is 20.00 mole% methyl alcohol, 60.0 mole% methyl acetate, and 20.0 mole% acetic acid.

What is the mass of a sample containing 45.0 kmol of methyl acetate?

Answer 1

4714.950 kilograms is the mass of a sample containing 45.0 kmol of methyl acetate.

Step-by-step explanation:

Moles of methyl acetate =
`n_1`=45.0 kmol= 45000 mol

Mole percentage of methyl acetate = 60.0%

Total moles in the sample = n

`60.0\%=(45000 mol)/(n)* 100`

`n=(45000 mol)/(60.0)* 100=75000 mol`

Mole percentage of methyl alcohol = 20.0%

Moles of methyl alcohol = n_2

`20.0\%=(n_2)/(75000 mol)* 100`

`n_2=15,000 mol`

Mass of methyl alcohol =
`n_2* 32.04 g/mol`

=
`15000 mol* 32.04 g/mol=480,600 g`

Mole percentage of acetic acid = 20.0%

Moles of acetic acid = n_3

`20.0\%=(n_3)/(75000 mol)* 100`

`n_3=15,000 mol`

Mass of acetic acid=
`n_3* 60.05 g/mol`

`15000 mol* 60.05 g/mol=900,750 g`

Mass of methyl methyl acetate=
`n_1* 74.08 g/mol`

`45000 mol* 74.08 g/mol =3,333,600 g`

Mass of sample: 480,600 g + 3,333,600 g + 900,750 g = 4714950 g

4714950 g = 4714.950 kg

(1 kg = 1000 g)