# 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?

## What is the total pressure, in atm, inside the flask?

Jul 17, 2016

2.05 mol.

#### Explanation:

First start by finding the mass of the gas mixture.

$\mathrm{de} n s i t y = \frac{m a s s}{v o l u m e}$

$m a s s = \mathrm{de} n s i t y \times v o l u m e$

$m a s s = 19.92 \setminus \frac{g}{L} \times 5.00 \setminus L$

$m a s s = 19.92 \setminus \frac{g}{\cancel{L}} \times 5.00 \setminus \cancel{L}$

$m a s s = 99.6 \setminus g$

The above mass is the mass of the gas mixture (${m}_{m i x}$). It includes the masses of ${N}_{2} , {H}_{2}$ and $C O$.

$\textcolor{red}{{m}_{m i x} = {m}_{{N}_{2}} + {m}_{{H}_{2}} + {m}_{C O}}$

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

underbrace(m_(N_2) = ???)

${m}_{{H}_{2}} = {n}_{{H}_{2}} \times M {M}_{{H}_{2}}$

${m}_{{H}_{2}} = 3.55 \setminus m o l . \times 2.016 \setminus \frac{g}{m o l .}$

${m}_{{H}_{2}} = 3.55 \setminus \cancel{m o l .} \times 2.016 \setminus \frac{g}{\cancel{m o l .}}$

$\underbrace{{m}_{{H}_{2}} = 7.16 \setminus g}$

${m}_{C O} = {n}_{C O} \times M {M}_{C O}$

${m}_{C O} = 1.25 \setminus m o l . \times 28.01 \frac{g}{m o l .}$

${m}_{C O} = 1.25 \setminus \cancel{m o l .} \times 28.011 \frac{g}{\cancel{m o l .}}$

$\underbrace{{m}_{C O} = 35.0 \setminus g}$

${m}_{{N}_{2}} = {m}_{m i x} - \left\{{m}_{{H}_{2}} + {m}_{C O}\right\}$

${m}_{{N}_{2}} = 99.6 \setminus g - \left\{7.16 \setminus g + 35.0 \setminus g\right\}$

$\underbrace{{m}_{{N}_{2}} = 57.4 \setminus g}$

Once the mass of ${N}_{2}$ is determined, find the number of moles.

${n}_{{N}_{2}} = \frac{{m}_{{N}_{2}}}{M {M}_{{N}_{2}}}$

${n}_{{N}_{2}} = \frac{57.4 \setminus g}{28.02 \setminus g . m o l {.}^{-} 1}$

${n}_{{N}_{2}} = \frac{57.4 \setminus \cancel{g}}{28.02 \setminus \cancel{g} . m o l {.}^{-} 1}$

${n}_{{N}_{2}} = 2.05 \setminus m o l .$