Answer:
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 ×[tex]\frac{1 moles O_{2}}{4 moles HCl}[/tex] = 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 ×[tex]\frac{2 moles Cl_{2}}{4 moles HCl}[/tex] = 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:
[tex]E = \frac{n_{f} - n_{i}}{m}[/tex]
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 × [tex]\frac{4 moles HCl}{2 moles Cl_{2}}[/tex] = 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!
