Question

A 2.47×10^-1 g sample of HCl and a 8.82 g sample of O₂ react in a closed 2.25 L container at 487 K, according to the following balanced chemical equation:

4HCl(g) + O₂(g) → 2H₂O(l) + 2Cl₂(g)

Calculate the P_Cl2 (in atm) in the container after the reaction has gone to completion.

Answer 1

Mass of HCl = 2.47*10^-1 g = 0.247 g
Molar mass of HCl = 36.5 g/mol
number of moles of HCl = mass / molar mass
                                                  =0.247/36.5
                                                  = 0.0068 mol

mass of O2 = 8.82 g
Molar mass of O2 = 32 g/mol
number of moles of O2 = mass / molar mass
                                                  =8.82/32
                                                  = 0.28 mol

4 mol of HCl require 1 mol of O2
Clearly, HCl is limiting eagent

We will use HCl in our further calculations

Now look at the equation:
4 mol of HCl forms 2 moles of Cl2
So, number of moles of Cl2 formed = 0.0068 mol / 2 = 0.0034 mol

So,
n = 0.0034 mol
V = 2.25 L
T = 487 K

usE:
P*V = n*R*T
P*2.25 = 0.0034 * 0.0821 * 487
P = 0.06 atm
Answer: PCl2 = 0.06 atm