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?

1 Answer
Jul 17, 2016

Answer:

2.05 mol.

Explanation:

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.#