Respuesta :
Answer:
- 1. mole fraction of hydrogen gas: 0.667
- 2. Mole fraction of oxygen gas: 0.333
- 3. Partial pressure of the hydrogen gas: 2.764 atm
- 4. Partial pressure of oxygen gas: 1.382 atm
Explanation:
1. Chemical equation of electrolysis
a) Word equation:
- Water → hydrogen gas + oxygen gas
b) Balanced chemical equation:
[tex]2H_2O(l)\rightarrow 2H_2(g)+O_2(g)[/tex]
c) Mole ratios:
[tex]2molH_2O(l):2molH_2(g):1molO_2(g)[/tex]
2. Number of moles of water
i) Formula: Number of moles = mass in grams/molar mass
ii) Molar mass of water: 18.015g/mol
iii) Mass: 27.1 g
iv) Substitute and compute:
- Number of moles = 27.1g / 18.015g = 1.504mol
3. Mol fraction of hydrogen gas:
a) Number of moles of hydrogen gas:
Using the theoretical mol ratio of hydrogen to water, 2 moles of water produce 2 moles of hydrogen gas. Then, 1.504 mol of water produce 1.504 mol of hydrogen gas.
- 1.504 mol H₂(g)
b) Number of moles of oxygen gas:
Using the theoretical mol ratio, 2 moles of water produce 1 mole of oxygen gas. The, 1.504 mol of water produce half of 1.504 mol = 0.752 moles of oxygen gas.
- 0.752 mol O₂(g)
c) Mole fraction: [tex]X_A[/tex]
i) Formula: [tex]X_A=\dfrac{\text{moles of A}}{\text{total number of moles}}[/tex]
ii) For hydrogen:
[tex]X_H_2=\dfrac{1.50mol}{1.504mol+0.725mol}[/tex]
[tex]X_H_2=0.667[/tex]
4. Mole fraction of oxygen gas
The mole fraction of all the components of a mixture adds up 1.
The mole fraction of O₂ = 1 - 0.667 = 0.333
5. Partial pressure of the hydrogen gas
i) Calculate the total pressure
- Use ideal gas equation:
[tex]pV=nRT[/tex]
[tex]p=\dfrac{nRT}{V}[/tex]
Substitute:
- n = 2.256mol
- R = 0.08206amt-liter/K-mol
- T = 273.15K
[tex]p=\dfrac{2.256atm\times 0.08206(atm\cdot liter/K\cdot mol)\times 273.15K }{12.2liter}\\\\\\p=4.146atm[/tex]
ii) Partial pressure
- Formula:
[tex]\text{Partial pressure of A}=X_A\times \text{Total pressure}[/tex]
- Substitute and compute:
[tex]\text{Partial pressure of }H_2=0.667\times 4.146atm=2.764atm[/tex]
6. Partial pressure oxygen gas
[tex]\text{Partial pressure of }O_2=0.333\times 4.146atm=1.382atm[/tex]
1. Mole fraction of hydrogen gas: 0.667
2. Mole fraction of oxygen gas: 0.333
3. Partial pressure of the hydrogen gas: 2.764 atm
4. Partial pressure of oxygen gas: 1.382 atm
Chemical equation of electrolysis:
a) Word equation:
Water → hydrogen gas + oxygen gas
b) Balanced chemical equation:
[tex]2H_2O--- > 2H_2(g)+O_2(g)[/tex]
Number of moles of water
Number of moles = mass in grams/molar mass
Molar mass of water: 18.015g/mol
Mass: 27.1 g
On substituting the values:
Number of moles = 27.1g / 18.015g = 1.504mol
Mol fraction of hydrogen gas:
a) Number of moles of hydrogen gas:
Using the theoretical mol ratio of hydrogen to water, 2 moles of water produce 2 moles of hydrogen gas. Then, 1.504 mol of water produce 1.504 mol of hydrogen gas.
1.504 mol H₂(g)
b) Number of moles of oxygen gas:
Using the theoretical mol ratio, 2 moles of water produce 1 mole of oxygen gas. The, 1.504 mol of water produce half of 1.504 mol = 0.752 moles of oxygen gas.
0.752 mol O₂(g)
c) Mole fraction:
For hydrogen: 1.50 mol / 1.504+0.725 mol
Mol fraction of hydrogen = 0.667
Mole fraction of oxygen gas
The mole fraction of all the components of a mixture adds up 1.
The mole fraction of O₂ = 1 - 0.667 = 0.333
Partial pressure of the hydrogen gas
i) Calculate the total pressure
Use ideal gas equation:
PV= nRT
where,
n = 2.256mol
R = 0.08206amt-liter/K-mol
T = 273.15K
P= (2.256 * 0.08206*273.15 )/ 12.2
P= 4.126 atm
Partial pressure of A = XA * Total pressure
Partial pressure of hydrogen gas = 0.667 * 4.146 = 2.764
Partial pressure oxygen gas = 0.333*4.146 = 1.382 atm
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