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Dominik [7]
3 years ago
11

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

Chemistry
1 answer:
svet-max [94.6K]3 years ago
5 0

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 <u>calculate the moles of HCl and then we use PV=nRT to calculate P</u>:

  • 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. <u>The total pressure is the sum of the pressures of all components</u>. Before the reaction took place, there are only reactants:

  • P = PHCl + PO₂
  • P = 0.071 atm + 0.018 atm = 0.089 atm

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

So now we <em>calculate the pressure of Cl₂ and of H₂O</em>:

  • 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 <u>total pressure</u> thus is:

  • P = PCl₂ + PH₂O
  • P = 0.035 atm + 0.035 atm = 0.070 atm
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