Question

A sample containing precious chemicals (MF and BAD) is given to you to analyze anddetermine the molar compositions. You can’t measure MF directly so you must send itthrough a reactor where it reacts with BAD in three reactions:

R1: MF + BAD→MAD + FB
R2: MF + 3BAD→B2 + F(AD)3 + MB R3: 2MF + 3BAD→2MBAD + 2FD+BA

The selectivity of FB to MB is 13.4 and the selectivity of B2 to BA is 165. If your equipment detects 2 moles of BA:

What is the molar composition of the original sample?

Can you determine the product stream composition? If so, what is it?

Answer 1

Part a

selectivity of FB to MB = 13.4

Moles of FB / moles of MB = 13.4.......... Eq1

selectivity of B2 to BA = 165

Moles of B2 / moles of BA = 165............ Eq2

moles of BA = 2 mol

In R3

Moles of moles of BAD consumed = 2 x 3 = 6 mol

Moles of MF consumed = 2 x 2 = 4 mol

From eq2

Moles of B2 = 165 x 2 = 330 mol

From the stoichiometry of the R2

Molar ratio of MB and B2 = 1 : 1

Moles of MB = moles of B2 = 330 mol

In R2

Moles of MF consumed = 330 mol

Moles of BAD consumed = 3 x 330 = 990 mol

From eq1

Moles of FB = 13.4 x 330 = 4422 mol

From the stoichiometry of the R1

1 mol MF produces = 1 mol FB

In R1

Moles of MF consumed = 4422 mol

Moles of BAD consumed = moles of MF = 4422 mol

Total Moles of MF consumed = 4422 + 330 + 4 = 4756 mol

Total Moles of BAD consumed = 4422 + 990 + 6 = 5418 mol

Total molar feed = 4756 + 5418 = 10174 mol

Molar composition of MF in original sample = 4756/10174

= 0.4675 = 46.75%

Molar composition of MF in original sample = 5418/10174

= 0.5325 = 53.25%

Part b

In R1

Moles of MAD = 4422 mol

Moles of FB = 4422 mol

In R2

Moles of B2 = 330 mol

Moles of F(AD)3 = 330 mol

Moles of MB = 330 mol

In R3

Moles of BA = 2 mol

Moles of MBAD = 2 x 2 = 4 mol

Moles of FD = 2 x 2 = 4 mol

Total Moles of product stream = 4422 + 4422 + 330 + 330 + 330 + 2 + 4 + 4

= 9844 mol

Mol% of MAD = 4422*100/9844 = 44.92 %

Mol% of FB = 4422*100/9844 = 44.92 %

Mol% of B2 = 330*100/9844 = 3.35 %

Mol% of F(AD)3 = 330*100/9844 = 3.35 %

Mol% of MB = 330*100/9844 = 3.35 %

mol% of BA = 2*100/9844 = 0.02%

mol% of MBAD = 4*100/9844 = 0.04%

mol% of FD = 4*100/9844 = 0.04%