Question

The following reaction was performed in a 5.70 L non-leaking container at a constant temperature of 106.8 ?C.

4HCl(g)+O2(g)?2Cl2(g)+2H2O(g)
Upon completion of the reaction 0.460 g Cl2 was formed.
1. What was the minimum starting pressure of HCl?
2. What was the minimum starting pressure of O2?
3. If stoichiometric amounts of HCl and O2 were used and the reaction went to completion, what was the total pressure of the system before the reaction took place?
4. If stoichiometric amounts of HCl and O2 were used, what was the total pressure of the system after the reaction went to completion?

Answer 1

1. 0.071 atm

2. 0.018 atm

3. 0.089 atm

4. 0.070 atm

Step-by-step explanation:

4HCl(g)+O₂(g) → 2Cl₂(g)+2H₂O(g)

1. First we calculate the moles of HCl and then we use PV=nRT to calculate P:

• 0.460gCl₂ ÷ 70.9g/1molCl₂ *
  `(4molHCl)/(2molCl_(2))` = 0.0130 mol HCl

• P * 5.70 L = 0.0130 mol * 0.082 atm·Lmol⁻¹·K⁻¹ * 379.96 K

• P = 0.071 atm

2. We do the same but with O₂

• 0.460gCl₂ ÷ 70.9g/1molCl₂ *
  `(1molO_(2))/(2molCl_(2))` = 3.244x10⁻³mol O₂

• P * 5.70 L =3.244x10⁻³mol * 0.082 atm·Lmol⁻¹·K⁻¹ * 379.96 K

• P = 0.018 atm

3. The total pressure is the sum of the pressures of all components. Before the reaction took place, there are only reactants:

• P = PHCl + PO₂

• P = 0.071 atm + 0.018 atm = 0.089 atm

4. After the reaction went to completion, there are only products, because stoichiometric amounts of reactants were used.

So now we calculate the pressure of Cl₂ and of H₂O:

• Cl₂ ⇒ 0.460gCl₂ ÷ 70.9g/1molCl₂ = 6.488x10⁻³ molCl₂

• P * 5.70 L =6.488x10⁻³mol * 0.082 atm·Lmol⁻¹·K⁻¹ * 379.96 K

P = 0.035 atm

• H₂O ⇒ 0.460gCl₂ ÷ 70.9g/1molCl₂ *
  `(2molH_(2)O)/(2molCl_(2))` = 6.488x10⁻³ molH₂O

• P * 5.70 L =6.488x10⁻³mol * 0.082 atm·Lmol⁻¹·K⁻¹ * 379.96 K

P = 0.035 atm

The total pressure thus is:

• P = PCl₂ + PH₂O

• P = 0.035 atm + 0.035 atm = 0.070 atm