Question

The partial pressure of CH4 is 0.175 atm and that of O2 is 0.250 atm in a mixture of the two gases. What is the mole fraction of each gas in the mixture? If the mixture occupies a volume of 10.5 L at 35oC, calculate the total number of moles of gas in the mixture. Calculate the number of grams of each gas in the mixture.

Answer 1

Answer:

The total number of moles of gas in the mixture is 0.16939.

1.25550 grams of methane gas and 3.18848 grams of oxygen gas.

Explanation:

Total volume of the mixture = V = 10.5 L

Temperature of the mixture = T = 35°C = 308.15K

Pressure of the mixture = P

Total moles of mixture = n =$n_1+n_2$

Using an ideal gas equation :

$PV=nRT$

$P\times 10.0 L=n\times 0.0821 atm L/mol K\times 308.15 K$

$P=2.509 n$

Partial pressure of the methane= $p_1=0.175 atm$

Moles of the methane= $n_1$

Partial pressure of the oxygen gas= $p_2=0.250 atm$

Moles of the methane= $n_2$

Mole fraction  of the methane= $\chi_1=\frac{n_1}{n_1+n_2}$

Mole fraction  of the oxygen gas= $\chi_2=\frac{n_2}{n_1+n_2}$

$p_1=p\times \chi_1$  (Dalton's law)

$0.175 atm=P\times \frac{n_1}{n_1+n_2}$

$0.175 atm=(2.509 n)\times \frac{n_1}{n}$

$n_1=0.06975 mol$

Mass of 0.06975 moles of methane gas :

0.06975 mol × 18 g/mol =1.25550 g

$p_2=p\times \chi_2$  (Dalton's law)

$0.250 atm=P\times \frac{n_2}{n_1+n_2}$

$0.250 atm=(2.509 n)\times \frac{n_2}{n}$

$n_2=0.09964 mol$

Mass of 0.06975 moles of oxygen gas :

0.09964 mol × 32 g/mol =3.18848 g

Total moles of mixture = n =$n_1+n_2$

[/tex]=0.06975 mol+0.09964 mol=0.16939 mol[/tex]