Q3-che-112p-4th-q-2015-1.docx

1. Cyclohexane (C6H12) can be made by the reaction of benzene (C6H6) with hydrogen to the following reaction C6H6 + 3H2 C6H12 The hydrogen and benzene in the fresh feed are maintained at a ratio of 3:1, same as in the theoretical ratio in the balance equation. (This makes the ratio of unreacted hydrogen and benzene also at 3:1) An overall conversion of 96 % was obtained when the recycle to fresh feed ratio is equal to 4. Calculate the conversion per passed (single pass conversion)

Basis: 100 moles of fresh Feed In fresh feed C6H6 = 25 mol H2 = 75 mol

1 𝑚𝑜𝑙𝑒 𝑐𝑦𝑐𝑙𝑜ℎ𝑒𝑥𝑎𝑛𝑒 1 𝑚𝑜𝑙 𝑏𝑒𝑛𝑧𝑒𝑛𝑒

C6H12 = 0.96(25)mole benzene x R/F = 4

Let x be the fractional conversion per pass R = 400 [25+ 0.25(400)] x = 24 X = 0.192 % conversion per pass = 19.2 %

= 24

2. Calculate the amount of secondary air used in a two-stage converter below using 100 moles of F as basis. What % of the SO2 is converted to SO3?

CG Basis: 100 moles of F S balance: 10 = [ 0.097 + 0.0053]CG CG = 95.2 mol N2 balance: 81 + 0.79 (2nd air) = 0.8504(95.2) 2nd air = -0.04

0

% conversion to SO3 = [ 0.0997(95.2) / 10] x 100 = 94.9 %

3. In the manufacture of sulfuric acid by the Contact Process, The feed is burned completely and analysis of the Burner gas shows 7.70% SO2, 11.02% O2 and 81.28% N2 Calculate the % of the Sulfur in the feed that is converted to SO3 if the feed to the Burner is a) Sulfur b) Pyrites a)

Air

Sulfur

Burner

7.70 % SO2 11.02 % O2 81.28 % N2 SO3

Basis: 100 moles of SO3-free Burner gas O2 supplied = 81.28(21/79) = 21.61 mol O2 in SO3 = 21.61 – 7.70 – 11.02 = 2.89 mol SO3 = 2.89(2/3) = 1.92 mol

%S

SO3 = [ 1.92 / (7.7 + 1.92)] x 100 = 19.95 %

b)

Air

FeS2

Burner

7.70 % SO2 11.02 % O2 81.28 % N2

Fe2O3 .

Basis: 100 moles of SO3-free Burner gas 2 FeS2 + 11/2 O2 Fe2O3 + 4 SO2 2 FeS2 + 15/2 O2 Fe2O3 + 4 SO3

O2 supplied = 81.28(21/79) = 21.61 mol O2 used in the first equation 7.70 (11/2 / 4) = 10.59 mol O2 used in the second equation = 21.61 – 10.59 - 11.02 = 0 mol Therefore, SO3 = 0 %S

SO3 = 0 %

SO3

4.

Calculate the % conversion of NH3

A, air P F,

NH3

Reactor

0.8 % NH3 9.5 % HNO3 3.8 % O2 85.9 % N2 x (mol) = H2O

Required: % conversion of ammonia Basis: 100 moles of H2O free product gases N balance F + 0.79 A(2) = 0.8 + 9.5 + 85.9(2)

O balance 0.21 A(2) = 9.5(3) + 3.8(2) +

x

H balance F(3) = 0.8(3) + 9.5 + x (2) X = 9.56 mol F = 10.34 mol A = 108.9 mol

% conversion of NH3 = ( fixed N2 in product / N2 in Feed NH3 ) x 100 =

𝑁2 𝑖𝑛 𝐻𝑁𝑂3 𝑁2 𝑖𝑛 𝑁𝐻3 9.5

x 100

= 10.34 x 100 = 91.9 %