Question

A gas mixture called heliox, 6.11% o2 and 93.89% he by mass, is used in scuba tanks for descents more than 65 m below the surface. calculate the mole fractions of he and o2 in this mixture. he is 0.9980

Answer 1

Answer: The mole fractions of He in the mixture is 0.992.

The mole fraction of $O_2$ in the mixture is 0.008.

Explanation:

A gas mixture called heliox, 6.11% $O_2$ and 93.89% He gas by mass by mass.

Percentage of oxygen gas in a mixture by mass = 6.11%

In 100 gram mixture 6.11% of oxygen is present by mass :$\frac{100\times 6.11}{100}=6.11 grams$

Percentage of helium gas in a mixture by mass = 93.89%

In 100 gram mixture 93.89% of helium is present by mass :$\frac{100\times 93.89}{100}=93.89 grams$

Mole fraction of component 'x' in a mixture of 'x' and 'y'= $=\chi _x=\frac{n_x}{n_x+n_y}$

$n_x$ = moles of gas 'x'=$\frac{\text{mass of gas 'x'}}{\text{Molar mass of gas 'x'}}$

$n_y$ = moles of gas 'y'=$\frac{\text{mass of gas 'y'}}{\text{Molar mass of gas 'y'}}$

Moles of oxygen gas ,when 6.11 grams of oxygen in present in a mixture:

$n_{O_2}=\frac{6.11 g}{16g/mol}=0.3818 moles$

Moles of helium gas ,when 93.89 grams of helium in present in a mixture:

$n_{He}=\frac{93.89 g}{2g/mol}=46.945 moles$

$=\chi _{O_2}=\frac{n_{O_2}}{n_{O_2}+n_{He}}=\frac{0.3818 mol}{0.3818 mol+46.945 mol}=0.008$

$=\chi _{He}=\frac{n_{He}}{n_{O_2}+n_{He}}=\frac{46.945 mol}{0.3818 mol+46.945 mol}=0.9919\approx 0.992$

The mole fractions of He in the mixture is 0.992.

The mole fraction of $O_2$ in the mixture is 0.008.

Answer 2

098.9 dats wat I think it is