3.55 moles $H_2$ (g), 1.25 moles $CO$ (g), and an unknown number of moles $N_2$ (g) were placed into a 5.00 L flask at $25.0^oC$. Given the density of this gas mixture to be 19.92 g/L, how many moles of $N_2$ were in the flask?

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Md Ariful Islam · Answer 1 · Apr 02, 2024 09:48:30

First start by finding the mass of the gas mixture.

$density = (mass )/ (volume)$

$mass = density xx volume$

$mass = 19.92 \ g/L xx 5.00 \ L$

$mass = 19.92 \ g/cancel(L)xx 5.00 \ cancel(L)$

$mass = 99.6 \ g$

The above mass is the mass of the gas mixture ($m_(mix)$). It includes the masses of $N_2 , H_2 $ and $CO$.

$color (red) (m_(mix) = m_(N_2) + m_(H_2) + m_(CO))$

$----------------$

$underbrace(m_(N_2) = ???)$

$m_(H_2) = n_(H_2) xx MM_(H_2)$

$m_(H_2) = 3.55 \ mol. xx 2.016 \ g/(mol.)$

$m_(H_2) = 3.55 \ cancel(mol.)xx 2.016 \ g/(cancel(mol.))$

$underbrace(m_(H_2) = 7.16 \ g)$

$m_(CO) = n_(CO) xx MM_(CO)$

$m_(CO) = 1.25 \ mol. xx 28.01 g/(mol.)$

$m_(CO) = 1.25 \ cancel(mol.) xx 28.011 g/(cancel(mol.))$

$underbrace (m_(CO) = 35.0 \ g)$

$m_(N_2) = m_(mix) -{ m_(H_2) + m_(CO)}$

$m_(N_2) = 99.6 \ g - { 7.16 \ g +35.0 \ g}$

$underbrace (m_(N_2) = 57.4 \ g)$

Once the mass of $N_2$ is determined, find the number of moles.

$n_(N_2) = (m_(N_2))/ (MM_(N_2))$

$n_(N_2) = (57.4 \ g)/ (28.02 \ g.mol.^-1)$

$n_(N_2) = (57.4 \ cancel(g))/ (28.02 \ cancel(g).mol.^-1)$

$n_(N_2) = 2.05 \ mol.$

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