In the process for the manufacture of chlorine by direct oxidation of hcl with air over a catalyst to form cl2 and h2o (only), the exit product is composed of mole fractions of cl2 (21.4%), h2o (21.4%), and n2 (45.0%) – the remainder is hcl and o2. calculate and /or identify: (a) the limiting reactant? (b) the percentage excess reactant? (c) the degree of completion of the reaction? (d) the extent of reaction?

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 × = 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 × = 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:

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 × = 0,001 moles of HCl. As initial moles are 0,429 and stoichiometric coefficient is 4, extent of reaction is 0,107

I hope it helps!