# Question c7d69

Apr 6, 2015

The mole fraction of each gas is $\text{0.208}$ for chlorine pentafluoride and $\text{0.793}$ for carbon dioxide.

The mole fraction of a gas, or $\chi$, that's part of a gaseous mixture is defined as the number of moles of that gas divided by the total number of moles in the mixture.

This means that, to get the mole fraction of each of the two gases, you must first determine how many moles of each you have, and then how many total moles you have in the mixture.

Use each of the two compounds' molar mass to determine how many moles of each you have

5.41cancel("g") * "1 mole ClF"_5/(130.445cancel("g")) = "0.04147 moles " $C l {F}_{5}$

6.97cancel("g") * "1 mole CO"_2/(44.01cancel("g")) = "0.1584 moles " $C {O}_{2}$

The total number of moles in the mixture will be

${n}_{\text{total}} = {n}_{C l {F}_{5}} + {n}_{C {O}_{2}}$

n_("total") = 0.04147 + 0.1584 = "0.19987 moles"#

Therefore,

${\chi}_{C l {F}_{5}} = \left(0.04147 \cancel{\text{moles"))/(0.19987cancel("moles}}\right) = 0.2074$

${\chi}_{C {O}_{2}} = \left(0.1584 \cancel{\text{moles"))/(0.19987cancel("moles}}\right) = 0.7925$

Rounded to three sig figs, the number of sig figs given for the mass of each gas, the answer wil be

${\chi}_{C l {F}_{5}} = \textcolor{g r e e n}{0.207}$
${\chi}_{C {O}_{2}} = \textcolor{g r e e n}{0.793}$