What volume of O_2 at 912 mmHg and 21° C is required to synthesize 18.0 mol of NO?

1 Answer
May 16, 2016

V_(O_2)=181L

Explanation:

Assuming the nitrogen monoxide is made from oxygen and nitrogen according to the following reaction:

N_2(g)+O_2(g)->2NO(g)

Therefore, 2 moles of NO are prepared using only 1 mole of O_2, thus, we will need 9.00 moles of O_2 to prepare 18.0 moles of NO.

The pressure of O_2 is P=912cancel(mmHg)xx(1atm)/(760cancel(mmHg)=1.20atm

The Kelvin temperature is T=21+273=294K

Therefore, using the ideal gas law, PV=nRT, the volume of oxygen can be found by:

V=(nRT)/P=(9.00cancel(mol)xx0.0821(cancel(atm)*L)/(cancel(mol)*cancel(K))xx294cancel(K))/(1.2cancel(atm))=181L