Question

In a process for the manufacture of chlorine by direct oxidation of HCl with air over a catalyst to form C12 and H2O (only), the exit product is composed of HCI (4.4%), CZ (19.8%), 02 (4,0%), and N2 (52.0%). What was? a) The limiting reactant? b) The percent excess reactant? The degree of completion of the reaction? d) The extent of reaction

Answer 1

a) The limiting reactant is HCl.

b) The percentage excess reactant is 88.5%

c) The degree of completion of the reaction is 99.8%

d) The extent of reaction is 0.107

Explanation:

The oxidation of HCl with O₂ is:

$4 HCl + O₂ → 2 Cl₂ + 2 H₂O$

In air, 78% is N₂ and O₂ is 21%

The initial mole fraction of gases is:

Y + X×0.78 + X×0.21 = 1

Where Y is the mole fraction of HCl and X is the mole fraction of air.

As Nitrogen doesn't react X×0.78 = 0.45. Thus,

X = 0.577.

$Y +( 0.577×0.78 )+ (0.577×0.21 )= 1\\ Y = 0.429$

a) For a complete reaction of 0,429 moles of HCl it is necessary:

0.429 moles HCl ×$\frac{1 moles O_{2}}{4 moles HCl}$ = 0,10725 moles of O₂

As you have 0.577×0.21 = 0.1212 moles of O₂

The limiting reactant is HCl.

b) The percentage excess reactant is:

0.1075 moles / 0.1212 moles × 100 = 88.5%

c) As the percent of products is 21.4% + 21.4% = 42.8%

0.429 moles of HCl produce × $\frac{2 moles Cl_{2}}{4 moles HCl}$ = 0,2145 moles of Cl₂ that are the same than moles of H₂O.

As mole percent of Cl₂ and H₂O is 21.4%. The degree of completion of the reaction is:

0.214/0.2145×100 = 99.8%

d) The extent of reaction is defined as:

$E = \frac{n_{f} - n_{i}}{m}$

Where n are the final and initial moles of reactant and m is the stoichiometric coefficient.

For HCl the final moles are:

0.429 - 0.214 moles of Cl₂ produce × $\frac{4 moles HCl}{2 moles Cl_{2}}$ = 0.001 moles of HCl.

As initial moles are 0.429 and stoichiometric coefficient is 4, the extent of the reaction is 0.107