Question

2. In manufacture of chlorine, feed containing hydrochloric acid gas and air are fed to the reactor. The product gases leaving the reactor are found to contain 13.2% HCl, 6.3% 02, 42.9% N2, 30% Cl2 and 7.6% H2O (by weight). Calculate:
i) The percent excess air used.
ii) The composition of gases entering reactor
iii) The degree of completion of oxidation (conversion)
(Hint: Start solving by change the mass flow rate to mol flow rate first.)

Answer 1

The balanced reaction

4 HCl + (O2 + 3.76 N2) = 2 Cl2 + 2 H2O + 3.76 N2

Let the mass of product gas = 100 g

Moles of HCl = mass/molecular weight

= 13.2g / 36.5g/mol

= 0.3616 mol

Moles of O2 = 6.3g / 32g/mol = 0.1968 mol

Moles of N2 = 42.9/28 = 1.532 mol

Total Moles of air = moles of O2 + moles of N2

= 0.1968 + 1.532 = 1.7288 mol

Moles of Cl2 = 30/71 = 0.4225 mol

Moles of H2O = 7.6/18 = 0.4222 mol

At inlet

Moles of HCl converted

= 4 mol HCl x 0.4225 mol Cl2 produced / 2 mol Cl2

= 0.8450 mol

Total Moles of HCl = Moles of HCl converted + Moles of HCl at outlet

= 0.8450 + 0.3616

= 1.2066 mol

Stoichiometric Moles of O2

= 1 mol O2 x 0.8450 moles of HCl converted / 4 mol HCl

= 0.2113 mol

Stoichiometric air required = 0.2113/0.21 = 1.006 mol

Actual moles of O2 supplied = Stoichiometric Moles of O2 + Moles of O2 at outlet

= 0.2113 + 0.1968 = 0.4081 mol

Moles of N2 supplied = 1.532 mol

Actual Moles of air supplied = moles of O2 + moles of N2

= 0.4081 + 1.532

= 1.94 moles

Part a

% excess air used

= ( actual air - stoichiometric air) *100 / (stoichiometric air)

= (1.94 - 1.006)*100/(1.006)

= 92.84 %

Part b

Total gases entering = moles of HCl + moles of air

= 1.207 + 1.94

= 3.147 mol

Part c

Conversion = (HCl converted) *100 / (HCl supplied)

= (0.8450)*100/(1.2066)

= 70.06 %