Question

A mixture contains 35.07 grams of carbon dioxide, 27.93 grams of water vapor, 12.64 grams of nitrogen, and 5.54 grams of helium. The total pressure of the system is 12 atm, What is the partial pressure of the helium?

A. 0.88 atm
B. 0.82 atm
C. 0.073 atm
D. 0.068 atm

Answer 1

The partial pressure of He = 3.97 atm

Step-by-step explanation:

Given:

Mass of CO2 = 35.07 g

Mass of H2O = 27.93 g

Mass of N2 = 12.64 g

Mass of He = 5.54 g

Total pressure P = 12 atm

To determine:

The partial pressure of He

Calculation:

Based on Dalton's law, the partial pressure of a gas can be expressed as a product of its mole fraction and the total pressure

`P(gas)=X(gas)*P(total)-----(1)`

where X(gas) = mole fraction

`X(gas)=(moles(gas))/(moles(total))----(2)`

`Moles(CO2)=(mass(CO2))/(mol.wt.(CO2))=(35.07g)/(44g/mol)=0.7970`

`Moles(H2O)=(mass(H2O))/(mol.wt.(H2O))=(27.93g)/(18g/mol)=1.552`

`Moles(N2)=(mass(N2))/(mol.wt.(N2))=(12.64g)/(28g/mol)=0.4514`

`Moles(He)=(mass(He))/(at.wt.(He))=(5.54g)/(4g/mol)=1.385`

Therefore:

Moles of He = 1.385

Total moles = 0.7970+1.552+0.4514+1.385 =4.188

Substituting the appropriate values in equation (1) gives:

`P(He)=X(He)*P(total)`

`P(He)=(1.385)/(4.188)*12 atm = 3.97 atm`

Answer 2

`\boxed{\text{3.6 atm}}`

Step-by-step explanation:

For this question, we must use Dalton's Law of Partial Pressures:

The partial pressure of a gas in a mixture of gases equals its mole fraction times the total pressure:

`p = \chi p_{\text{tot}}`

1. Calculate the number of moles of each gas.

`n_{\text{CO}_(2)} = \text{35.07 g} * \frac{\text{1 mol}}{\text{44.01 g}} = \text{0.7969 mol}\\\\n_{\text{H}_(2)\text{O}} = \text{27.93 g} * \frac{\text{1 mol}}{\text{18.02 g}} = \text{1.550 mol}\\\\n_{\text{N}_(2)} = \text{12.64 g} * \frac{\text{1 mol}}{\text{14.01 g}} = \text{0.9022 mol}\\\\n_{\text{He}} = \text{5.54 g} * \frac{\text{1 mol}}{\text{4.003 g}} = \text{1.384 mol}`

2. Calculate the total moles

`n_{\text{tot}} = \text{(0.7969 + 1.550 + 0.9022 + 1.384) mol = 4.633 mol}`

3. Calculate the mole fraction of helium

`\chi = \frac{n_\text{He}}{n_{\text{tot}}} = \frac{\text{1.384 mol}}{\text{4.633 mol}}= 0.2987`

4. Calculate the partial pressure of helium:

`p_{\text{He}} = \chi_{\text{He}} p_{\text{tot}}= 0.2987 * \text{12 atm} = \textbf{3.6 atm}\\\\p_{\text{He}} = \boxed{\textbf{3.6 atm}}`