Question

In the formation of smog, nitrogen and oxygen gas react to form nitrogen dioxide:
N2(g)+2O2(g)→2NO2(g)

How many grams of NO2 will be produced when 2.1 L of nitrogen at 860 mmHg and 24 ∘C are completely reacted?

Answer 1

Solution-

Using the ideal gas law to calculate moles of N2:

PV = n RT

P = 860/760

=1.131 atm

Now the V (volume) = 2.1 L

T = 24+273 = 297 K

(1.131)(2.1) = n (0.0821) (297)

n = 0.0974 mol of N2

Now let's calculate moles of NO2 formed

0.0974 mol of N2 * (2 mol NO2/1 mol N2)

= 0.1948 mol NO2

Finally let's calculate the mass of NO2

0.1948 mol NO2 * 46.0 g/mol

= 8.96 grams of NO2