Question

Question 1
A solid hydrocarbon is burned in air in a closed container, producing a mixture of gases having a total
pressure of 3.34 atm. Analysis of the mixture shows it to contain 0.340 g of water vapor, 0.792 g of
carbon dioxide, 0.288 g of oxygen, 3.790 g of nitrogen, and no other gases. Calculate the mole fraction
and partial pressure of carbon dioxide in this​

Answer 1

Answer:

particles of gas and electricity bill dioxide emissions

Answer 2

Answer:

The mole fraction of carbon dioxide, $x_{CO_{2} }$, measured is $0.1$.

The partial pressure of carbon dioxide, $P_{CO_{2} }$, measured is $0.334atm$.

Explanation:

Data given,

In the given mixture:

The mass of water vapour, $m_{H_{2}O }$ = $0.340g$

The mass of carbon dioxide gas, $m_{CO_{2} }$ = $0.792g$

The mass of oxygen gas, $m_{O_{2} }$ = $0.288g$

The mass of nitrogen gas, $m_{N_{2} }$ = $3.790g$

The total pressure of the mixture of gases, P = $3.34atm$

The mole fraction of carbon dioxide, $x_{CO_{2} }$ =?

The partial pressure of carbon dioxide, $P_{CO_{2} }$ =?

As we know,

• Mole fraction = $\frac{n}{N}$

Here, n = The number of moles of individual gas

And N = The total number of moles

Also,

• Number of moles = $\frac{Given mass}{Molar mass}$

1) For water vapour,

The molar mass of water ( $H_{2}O$ ) = $18g/mol$

• $n_{H_{2}O }$ = $\frac{0.340}{18}$ = $0.018mol$

2) For carbon dioxide gas,

The molar mass of carbon dioxide ( $CO_{2}$ ) = $44g/mol$

• $n_{CO_{2} }$ = $\frac{0.792}{44}$ = $0.018mol$  

3)  For oxygen gas,

The molar mass of oxygen gas ( $O_{2}$ ) = $32g/mol$

• $n_{O_{2} }$ = $\frac{0.288}{32}$ = $0.009mol$  

4)  For nitrogen gas,

The molar mass of nitrogen gas ( $N_{2}$ ) = $28g/mol$

• $n_{N_{2} }$ = $\frac{3.790}{28}$ = $0.135mol$  

Now, the total number of moles = $n_{H_{2}O } + n_{CO_{2} } + n_{O_{2} } + n_{N_{2} }$

• The total number of moles, N = $0.018 + 0.018 + 0.009 + 0.135$ = $0.18mol$

Thus,

• The mole fraction of carbon dioxide, $x_{CO_{2} }$ = $\frac{n_{CO_{2} }}{N}$ = $\frac{0.018}{0.18}$ = $0.1$.

And,

• The partial pressure of carbon dioxide, $P_{CO_{2} }$ = $x_{CO_{2} } \times P$ = $0.1 \times 3.34$ = $0.334atm$.