Two sequential hydrogenation reactions take place in a continuous reactor at ste

Question

Two sequential hydrogenation reactions take place in a continuous reactor at steady state.

C2H2+H2->C2H4

C2H4+H2->C2H6

The feed stream contains 17.0 mole% C2H2, 66.0 mole% H2 and 17.0 mole% inerts. The fractional conversion of C2H2 to C2H4 is 0.53 and the fractional conversion of C2H4 to C2H6 is 0.43

If the basis of the feed stream is 100 mol, what are the moles of each gas in the product stream?

1. number of mol c2h4

2.number of mol of c2h6

3.number of mol of H2

4.

What is the total moles of gas in the product stream?

5.

What is the percent by mole of each gas in the product stream?

1. % C2H2

2. % C2H4

3. % C2H6

4. % H2

5. % inerts

Explanation / Answer

If total feed is 100 mol
Amount of C2H2 in feed = 17 mol
Amount of H2 in feed = 66 mol

0.53 part of C2H2 converts to C2H4

so C2H4 formed = 0.53*17 = 9.01 mol
C2H2 remaining = 17-9.01 =7.99 mol

0.43 part of C2H4 forms C2H6
so C2H6 formed   = 0.43*9.01 = 3.87 mol

C2H4 remaining = 9.01 mol - 3.87 mol =5.14 mol

Total H2 used = 9.01 + 3.87 = 12.88 mol
H2 remaining = 66 - 12.88 =53.12 mol

Answer 1:
Mol of C2H4   = 5.14 mol

Answer2:
Mol of C2H6 = 3.87 mol

Answer 3:
Total H2 used = 9.01 + 3.87 = 12.88 mol
H2 remaining = 66 - 12.88 =53.12 mol
Mol of H2 = 53.12 mol

Answer 4:
Total mol of gas in product = Mol of C2H4 + Mol of C2H6
                                                        = 5.14 + 3.87
                                                         =9.01 mol

Answer 5:
Since total number of moles = 100
percent of C2H2 =number of moles of C2H2 *100/total moles
      = number of moles of C2H2 *100/total moles
       = number of moles of C2H2
   % C2H2 = 7.99 %
% C2H4 =5.14 %
%C2H6 = 3.87%
% H2 = 53.12 %
% inert gas = 17%

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