Question

The total pressure of a gaseous mixture of 2.8 g N2, 3.2 g O2and 0.5 g H2 is 4.5 atm. Calculate
the partial pressure of each gas.
A) P 1.00 atm; P :1.00 atm; P 2.5 atm
B) P 1.00 atm; P 200 atm; P 3.5 atm
©) Prz 2.00 atm; Poz:1.50 atm; Pre: 2.5 atm
D) P. 1.00 atm; P. :1.00 atm: P - 3.5 atm​

Answer 1

In your case, the total number of moles will be

ntotal=0.75+0.30+0.15=1.2 moles

This means that you have

PN2=(0.75moles)/(1.2moles)×1.56 atm=0.98 atm

PO2=(0.30moles)/(1.2moles)×1.56 atm=0.39 atm

PCO2=(0.15moles)/(1.2moles)×1.56 atm=0.20 atm

The values don';t add up to give 1.56 atm

because they must be rounded to two sig figs, the number of sig figs you have for the moles of each gas.

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Answer 2

Answer:

The partial pressure of the three gases, i.e., N₂, O₂, and, H₂ in the mixture whose total pressure is 4.5 atm are 1.035 atm, 1.035 atm, and, 2.52 atm respectively.
Thus, OPTION-A is the best choice here.

Explanation:

Dalton's law of Partial Pressure:

"The total pressure exerted by a mixture of gases is equal to the sum of the partial pressures of each of the individual gas".

Mathematically, Dalton's law can be represented as:

$P_{A} = \chi_{A} \times P_{total}$
where $P_{A}$ is the partial pressure of gas A in a mixture of gases which exert the total pressure of $P_{total}$.
$\chi_{A}$ = mole fraction of gas A in the mixture.

Calculation:

The total number of moles of each gas present in the mixture:

For nitrogen:
$n_{N_{2} } = \frac{2.8}{28} = 0.1$

For oxygen:

$n_{O_{2} } = \frac{3.2}{32} = 0.1$

For hydrogen:
$n_{H_{2} } = \frac{0.5}{2} = 0.25$

The mole fraction of each gas present in the mixture:

For nitrogen:
$\chi_{n_{N_{2}}} = \frac{n_{N_{2}}}{n_{N_{2}} + n_{O_{2}} +n_{H_{2}}} \\ \Rightarrow \chi_{n_{N_{2}}} = \frac{0.1}{0.1 + 0.1 + 0.25} \\\Rightarrow \chi_{n_{N_{2}}} = \frac{0.1}{0.45} \\\Rightarrow \chi_{n_{N_{2}}} = 0.2222 \approx 0.23$

For oxygen:
$\chi_{n_{O_{2}}} = \frac{n_{O_{2}}}{n_{N_{2}} + n_{O_{2}} +n_{H_{2}}} \\\Rightarrow \chi_{n_{O_{2}}} = \frac{0.1}{0.1 + 0.1 + 0.25} \\\Rightarrow \chi_{n_{O_{2}}} = \frac{0.1}{0.45} \\\Rightarrow \chi_{n_{O_{2}}} = 0.2222 \approx 0.23$

For hydrogen:
$\chi_{n_{H_{2}}} = \frac{n_{H_{2}}}{n_{N_{2}} + n_{O_{2}} +n_{H_{2}}} \\\Rightarrow \chi_{n_{H_{2}}} = \frac{0.25}{0.1 + 0.1 + 0.25} \\\Rightarrow \chi_{n_{H_{2}}} = \frac{0.25}{0.45} \\\Rightarrow \chi_{n_{H_{2}}} = 0.55556 \approx 0.56$

Calculation of partial pressure of each gas present in the mixture:

For nitrogen:
$P_{N_{2} } = \chi_{N_{2}} \times P_{total} \\\\\Rightarrow P_{N_{2} } = 0.23 \times 4.5 \\\\\Rightarrow P_{N_{2} } = 1.035 atm$

For oxygen:
$P_{O_{2} } = \chi_{O_{2}} \times P_{total} \\\\\Rightarrow P_{O_{2} } = 0.23 \times 4.5 \\\\\Rightarrow P_{O_{2} } = 1.035 atm$

For hydrogen:
$P_{H_{2} } = \chi_{H_{2}} \times P_{total} \\\\\Rightarrow P_{H_{2} } = 0.56 \times 4.5 \\\\\Rightarrow P_{N_{2} } = 2.52 atm$

For more information on Dalton's law on partial pressure, you may refer to the links below:

State Dalton's law of partial pressure.
https://brainly.in/question/2603827

state and write mathematical expression and Dalton law of partial pressure and explain it with suitable example​.
https://brainly.in/question/14404404

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