Question

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

Answer 1

Answer:

1. 0.071 atm

2. 0.018 atm

3. 0.089 atm

4. 0.070 atm

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 * $\frac{4molHCl}{2molCl_{2}}$ = 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 * $\frac{1molO_{2}}{2molCl_{2}}$ = 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  * $\frac{2molH_{2}O}{2molCl_{2}}$ = 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