Question

A process is carried out in which a mixture containing 25.0 wt% methanol, 42.5% ethanol, and the

balance water is separated into two fractions. A technician draws and analyzes samples of both product

streams and reports that one stream contains 39.8% methanol and 31.5% ethanol and the other contains

19.7% methanol and 41.2% ethanol. You examine the reported figures and tell the technician that they

must be wrong and that stream analyses should be carried out again. Is your statement correct?

Answer 1

Let moles of total sample= 1

This divided into two streams. Let the two stream be S1 and S2.

S1+S2= 1

Given moles of methanol in the sample= 0.25, ethanol= 0.425 and rest water= 1-0.25-0.425=0.325

Stream-1 : methanol= S1*0.398, ethanol = S1*0.315, water = S1*(1-0.398-0.315)=S1*0.287

Stream 2 : Methanol =S2*0.197, ethanol = S2*0.412, water =S2*(1-0.197-0.412)= S2*0.391

Writing methanol balance, 0.25= S1*0.398 +S2*0.197 or 0.25/0.398= S1+0.197/0.398

0.63= S1+0.5S2 (1), Writing ethanol balance 0.425= S1*0.315 +S2*0.412,

0.425/0.315= S1+S2*0.412/0.315, 1.35= S1+S2*1.31 (2)

Eq.2-Eq.1 gives S2*(1.31-0.63)= S2*(1.31-0.5) , S2= 0.84, S1= 1-0.84=0.16

Accordingly water in stream-1 = 0.16*0.287 = 0.046, water in S2= 0.84*0.391= 0.33

Total water= 0.33+0.046= 0.376 which is more than the water entering the system. Hence the analysis is incorrect.