How many grams of NO2

In the formation of smog, nitrogen and oxygen gas react to form nitrogen dioxide:
N2+2O2--->2NO2

How many grams of NO2 will be produced when 2.1 L of nitrogen at 820mmHg and 25 degrees celcius are completely reacted?

Step 1: Use the ideal gas law to calculate moles of N2:

PV = n RT
P = 820/760 = 1.08 atm
V = 2.1 L
T = 25+273 = 298 K

(1.08)(2.1) = n (0.0821) (298)
n = 0.0927 mol N2

2. Calculate moles of NO2 formed:

0.0927 mol N2 X (2 mol NO2/1 mol N2) = 0.185 mol NO2

3. Calculate mass of NO2:
0.185 mol NO2 X 46.0 g/mol = 8.53 grams NO2

first let us find how much nitrogen we have
PV=nRT
n=PV/RT where R = 62.36367 L mmHg K−1   mol−1
n=820*2.1/62.36367*(273+25)
n=0.09266 mole
one mole N2 = 28g therefor
28*0.09266 = 2.59g
From the balanced reaction equation
28g nitrogen produce 92g NO2 therefor
92*2.59/28 = 8.51g will be produced