3.55 moles H_2H2 (g), 1.25 moles COCO (g), and an unknown number of moles N_2N2 (g) were placed into a 5.00 L flask at 25.0^oC25.0oC. Given the density of this gas mixture to be 19.92 g/L, how many moles of N_2N2 were in the flask?

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

1 Answer
Jul 17, 2016

2.05 mol.

Explanation:

First start by finding the mass of the gas mixture.

density = (mass )/ (volume)density=massvolume

mass = density xx volumemass=density×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.