Question

A 8.00 L tank at 26.9 C is filled with 5.53 g of dinitrogen difluoride gas and 17.3 g of sulfur hexafluoride gas. You can assume both gases behave as ideal gases under these conditions. Calculate the mole fraction of each gas. Round each of your answers to significant digits.

Answer 1

Answer:

$x_{N_2F_2}= 0.415\\\\x_{SF_6}=0.585$

Explanation:

Hello!

In this case, since the mole fraction of both gases in the tank is computed via:

$x_{N_2F_2}=\frac{n_{N_2F_2}}{n_{N_2F_2}+n_{SF_6}} \\\\x_{SF_6}=\frac{n_{SF_6}}{n_{N_2F_2}+n_{SF_6}}$

It means we need to compute the moles of each gas, just as it is shown down below:

$n_{N_2F_2}}=5.53gN_2F_2*\frac{1molN_2F_2}{66.01gN_2F_2} =0.0838molN_2F_2\\\\n_{SF_6}=17.3gSF_6*\frac{1molSF_6}{146.06gSF_6} =0.118molSF_6$

Thus, the mole fractions turn out:

$x_{N_2F_2}=\frac{0.0838mol}{0.0838mol+0.118mol}= 0.415\\\\x_{SF_6}=\frac{0.0838mol}{0.0838mol+0.0838mol}=0.585$

Best regards!