Question

imagine that unknown gas in the storage tank is analyzed. a 28.05 - g sample of the gas contains 9.26 g c₂h₆, 8.83 g so₂, and 9.96 g o₂. calculate the partial pressures of each gas (in bar) in the storage tank if the total pressure in the tank is 2.45 bar.
a) pc₂h₆ 1.02 bar
b) pso₂ 0.455 bar
c) po₂ 1.03 bar
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Explanation:

Step1: Calculate moles of each gas

The molar mass of $C_2H_6$ is $M_{C_2H_6}=(2\times12.01 + 6\times1.01)\text{ g/mol}=30.07\text{ g/mol}$. The number of moles of $C_2H_6$, $n_{C_2H_6}=\frac{9.26\text{ g}}{30.07\text{ g/mol}} = 0.308\text{ mol}$.
The molar mass of $SO_2$ is $M_{SO_2}=(32.07+2\times16.00)\text{ g/mol}=64.07\text{ g/mol}$. The number of moles of $SO_2$, $n_{SO_2}=\frac{8.83\text{ g}}{64.07\text{ g/mol}}=0.138\text{ mol}$.
The molar mass of $O_2$ is $M_{O_2}=2\times16.00\text{ g/mol} = 32.00\text{ g/mol}$. The number of moles of $O_2$, $n_{O_2}=\frac{9.96\text{ g}}{32.00\text{ g/mol}}=0.311\text{ mol}$.

Step2: Calculate mole - fractions

The total number of moles, $n_{total}=n_{C_2H_6}+n_{SO_2}+n_{O_2}=0.308 + 0.138+0.311=0.757\text{ mol}$.
The mole - fraction of $C_2H_6$, $x_{C_2H_6}=\frac{n_{C_2H_6}}{n_{total}}=\frac{0.308\text{ mol}}{0.757\text{ mol}} = 0.407$.
The mole - fraction of $SO_2$, $x_{SO_2}=\frac{n_{SO_2}}{n_{total}}=\frac{0.138\text{ mol}}{0.757\text{ mol}}=0.182$.
The mole - fraction of $O_2$, $x_{O_2}=\frac{n_{O_2}}{n_{total}}=\frac{0.311\text{ mol}}{0.757\text{ mol}}=0.411$.

Step3: Calculate partial pressures

According to Dalton's law of partial pressures, $p_i = x_i\times p_{total}$. Given $p_{total}=2.45\text{ bar}$.
For $C_2H_6$, $p_{C_2H_6}=x_{C_2H_6}\times p_{total}=0.407\times2.45\text{ bar}=0.997\text{ bar}\approx1.00\text{ bar}$.
For $SO_2$, $p_{SO_2}=x_{SO_2}\times p_{total}=0.182\times2.45\text{ bar}=0.446\text{ bar}$.
For $O_2$, $p_{O_2}=x_{O_2}\times p_{total}=0.411\times2.45\text{ bar}=1.007\text{ bar}\approx1.01\text{ bar}$.

Answer:

a) $p_{C_2H_6}=1.00\text{ bar}$
b) $p_{SO_2}=0.446\text{ bar}$
c) $p_{O_2}=1.01\text{ bar}$