Question

A gas mixture consists of 4 kg of O2, 5 kg of N2, and 7 kg of CO2. Determine (a) the mass fraction of each component, (b) the mole fraction of each component, and (c) the average molar mass and gas constant of the mixture.

Answer 1

Answer:

See explanation.

Explanation:

Hello

(a) In this case, one uses the following formulas, which allow to compute the mass fraction of each component:$\% O_2=\frac{m_{O_2}}{m_{O_2}+m_{N_2}+m_{CO_2}}*100\%=\frac{4kg}{4kg+5kg+7kg}*100\%=25\%O_2\\\% N_2=\frac{m_{N_2}}{m_{O_2}+m_{N_2}+m_{CO_2}}*100\%=\frac{5kg}{4kg+5kg+7kg}*100\%=31.25\%N_2\\\% CO_2=\frac{m_{CO_2}}{m_{O_2}+m_{N_2}+m_{CO_2}}*100\%=\frac{7kg}{4kg+5kg+7kg}*100\%=43.75\%CO_2$

(b) For the mole fractions, it is necessary to find all the components' moles by using their molar mass as shown below:

$n_{O_2}=4kgO_2*\frac{1kmolO_2}{32kgO_2} =0.125kmolO_2\\n_{N_2}=5kgN_2*\frac{1kmolN_2}{28kgN_2} =0.179kmolN_2\\n_{CO_2}=7kgCO_2*\frac{1kmolCO_2}{44kgCO_2} =0.159kmolCO_2$

Now, the mole fractions:

$x_{O_2}=\frac{n_{O_2}}{n_{O_2}+n_{N_2}+n_{CO_2}}*100\%=\frac{0.125kmolO_2}{0.125kmolO_2+0.179kmolN_2+0.159kmolCO_2}*100\%=27\%O_2\\x_{N_2}=\frac{n_{N_2}}{n_{O_2}+n_{N_2}+n_{CO_2}}*100\%=\frac{0.179kmolN_2}{0.125kmolO_2+0.179kmolN_2+0.159kmolCO_2}*100\%=38.7\%N_2\\ x_{CO_2}=\frac{n_{CO_2}}{n_{O_2}+n_{N_2}+n_{CO_2}}*100\%=\frac{0.159kmolCO_2}{0.125kmolO_2+0.179kmolN_2+0.159kmolCO_2}*100\%=34.3\%CO_2$

(c) Finally the average molar mass is computed considering the molar fractions and each component's molar mass:

$M_{average}=0.27*32g/mol+0.387*28g/mol+0.343*44g/mol=34.57g/mol$

And the gas constant:

$Rg=0.082\frac{atm*L}{mol*K}*\frac{1mol}{34.57g}=0.00237\frac{atm*L}{g*K}$

Best regards.