Question

A mixture of A and B is capable of being ignited only if the mole percent of A is 6 %. A mixture containing 9.0 mole% A in B flowing at a rate of 800 kg/h is to be diluted with pure B to reduce A concentration to the lower flammability limit. Calculate the required flow rate of B in mol/h and the percent by mass of Oz in the product gas. (Note: B may be taken to consist of 21 mole% O2 and 79% Nz and to have an average molecular weight of 29.0.) MA = 16.04 g/mol.

Answer 1

Step-by-step explanation:

As it is given that mixture (contains 9 mol % A and 91% B) and it is flowing at a rate of 800 kg/h.

Hence, calculate the molecular weight of the mixture as follows.

Weight =
`0.09 * 16.04 + 0.91 * 29`

= 27.8336 g/mol

And, molar flow rate of air and mixture is calculated as follows.

`(800)/(27.8336)`

= 28.74 kmol/hr

Now, applying component balance as follows.

`0.09 * 28.74 + 0 * F_(B) = 0.06F_(p)`

`F_(p)` = 43.11 kmol/hr

`F_(A) + F_(B) = F_(p)`

`F_(B)` = 43.11 - 28.74

= 14.37 kmol/hr

So, mass flow rate of pure (B), is
`F_(B)` =
`14.37 * 29`

= 416.73 kg/hr

According to the product stream, 6 mol% A and 94 mol% B is there.

Molecular weight of product stream =
`Mol. weight * 43.11 kmol/hr`

=
`0.06 * 16.04 + 0.94 * 29`

= 28.22 g/mol

Mass of product stream = 1216.67 kg/hr

Hence, mole of
`O_(2)` into the product stream is as follows.

`0.21 * 0.94 * 43.11`

=
`8.5099 kmol/hr * 329 g/mol`

= 272.31 kg/hr

Therefore, calculate the mass % of
`O_(2)` into the stream as follows.

`(272.31)/(1216.67) * 100`

= 22.38%

Thus, we can conclude that the required flow rate of B is 272.31 kg/hr and the percent by mass of
`O_(2)` in the product gas is 22.38%.