Chemistry, 06.12.2019 21:31 bhadd2001

Amixture of helium and methane gases, at a total pressure of 821 mm hg, contains 0.723 grams of helium and 3.43 grams of methane. what is the partial pressure of each gas in the mixture?

phe = mm hg
pch4 = mm hg

2.) a mixture of nitrogen and carbon dioxide gases contains nitrogen at a partial pressure of 363 mm hg and carbon dioxide at a partial pressure of 564 mm hg. what is the mole fraction of each gas in the mixture?

xn2 =
xco2 =

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Answer from: melissa3333

For 1: The partial pressure of helium is 376 mmHg and that of methane gas is 445 mmHg

For 2: The mole fraction of nitrogen gas is 0.392 and that of carbon dioxide gas is 0.608

Explanation:

For 1:

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ .....(1)

For helium:

Given mass of helium = 0.723 g

Molar mass of helium = 4 g/mol

Putting values in equation 1, we get:

$\text{Moles of helium}=\frac{0.723g}{4g/mol}=0.181mol$

For methane gas:

Given mass of methane gas = 3.43 g

Molar mass of methane gas = 16 g/mol

Putting values in equation 1, we get:

$\text{Moles of methane gas}=\frac{3.43g}{16g/mol}=0.214mol$

To calculate the mole fraction , we use the equation:

$\chi_A=\frac{n_A}{n_A+n_B}$ .......(2)

To calculate the partial pressure of gas, we use the equation given by Raoult's law, which is:

$p_{A}=p_T\times \chi_{A}$ ......(3)

For Helium gas:

We are given:

$n_{He}=0.181mol\\n_{CH_4}=0.214mol$

Putting values in equation 2, we get:

$\chi_{He}=\frac{0.181}{0.181+0.214}=0.458$

Calculating the partial pressure by using equation 3, we get:

$p_T=821mmHg\\\\\chi_{He}=0.458$

Putting values in equation 3, we get:

$p_{He}=0.458\times 821mmHg=376mmHg$

For Methane gas:

We are given:

$n_{He}=0.181mol\\n_{CH_4}=0.214mol$

Putting values in equation 2, we get:

$\chi_{CH_4}=\frac{0.214}{0.181+0.214}=0.542$

Calculating the partial pressure by using equation 3, we get:

$p_T=821mmHg\\\\\chi_{CH_4}=0.542$

Putting values in equation 3, we get:

$p_{CH_4}=0.542\times 821mmHg=445mmHg$

Hence, the partial pressure of helium is 376 mmHg and that of methane gas is 445 mmHg

For 2:

We are given:

Partial pressure of nitrogen gas = 363 mmHg

Partial pressure of carbon dioxide gas = 564 mmHg

Total pressure = (363 + 564) mmHg = 927 mmHg

Calculating the mole fraction of the gases by using equation 3:

For nitrogen gas:

$363=\chi_{N_2}\times 927\\\\\chi_{N_2}=\frac{363}{927}=0.392$

For carbon dioxide gas:

$564=\chi_{CO_2}\times 927\\\\\chi_{CO_2}=\frac{564}{927}=0.608$

Hence, the mole fraction of nitrogen gas is 0.392 and that of carbon dioxide gas is 0.608

Answer from: Quest

answer : third trial had the contaminated sample and the correct value for dhrxn should be 1000 kj/mol

explanation :

let us write down the given equations

equation 1 : $4x (s) + 3o_{2} (g) \rightarrow 2x_{2}o_{3} (s) h rxn = -600 kj/mol$

equation 2 : $2x (s) + 3cl_{2} (g) \rightarrow 2xcl_{3} (s) h rxn = -800 kj/mol$

equation 3 : $4xcl_{3} (s) + 3o_{2} (g) \rightarrow 2x_{2}o_{3} (s) + 6cl_(2} h rxn = -200 kj/mol$

let us use hess's law to find out which 2 equations can be added to get the third equation.

let us reverse equation 2 and multiply it by 2. the new equation that we get is,

equation 4 : $4xcl_{3} (s) \rightarrow 4x (s) + 6cl_{2} ( h rxn = (+800) \times 2 kj/mol$

let us add equation 1 & 4

$4x (s) + 3o_{2} (g) \rightarrow 2x_{2}o_{3} (s) h rxn = -600 kj/mol$

$4xcl_{3} (s) \rightarrow 4x (s) + 6cl_{2} ( h rxn = (+1600) kj/mol$

$4xcl_{3} )s) + 3o_{2} \rightarrow 2x_{2}o_{3} (s) + 6cl_{2} (g) h rxn = (+1000) kj/mol$

the above equation is similar to equation 3 but has different value for dhrxn.

therefore third trial had the contaminated sample and the correct value for dhrxn should be 1000 kj/mol

Answer from: Quest

answersilicon (si); oxygen (o);

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Amixture of helium and methane gases, at a total pressure of 821 mm hg, contains 0.723 grams of heli...

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