Question

7 kg of dimethyl terephthalate is reacted with 5 kg of Ethylene glycol to produce PET. Estimate the amount of polymer that is produced assuming 50% efficiency.

Answer 1

Step-by-step explanation:

The given data will be as follows.

7 kg of dimethyl terephthalate + 5 kg of Ethylene glycol = 12 kg of product

Formula for weight average number is as follows.

`w_(n) = \sum w_(i) * M_(i)`

=
`(7)/(12) * M_(DMT) + (5)/(12) * M_(EG)`

Molecular weight of DMT = 198.21 g/mol

Molecular weight of ET = 62.07 g/mol

Hence, calculate the weight average number as follows.

`(7)/(12) * M_(DMT) + (5)/(12) * M_(EG)`

=
`(7)/(12) * 198.21 g/mol + (5)/(12) * 62.07 g/mol`

= (115.62 + 25.862) g/mol

= 141.482 g/mol

Since, efficiency is given as 50% = 0.5

0.5 =
`\frac{\bar{M_(avg. PET)}}{w_(n)}`

`\bar{M_(avg. PET)} = 0.5 * w_(n)`

=
`0.5 * 141.482 g/mol`

= 70.741 g/mol

As,
`m_(PET) * \bar{M_(avg. PET)}` =
`m_(DMT) * M_(DMT) + m_(EG) * M_(EG)`

`m_(PET) * 70.741 g/mol = 7 * 198.21 g/mol + 5 * 62.07 g/mol`

`m_(PET)` =
`(1697.82)/(70.741)`

= 24 g

Thus, we can conclude that the amount of given polymer is 24 g.