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

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Answer 1 · 2016-05-16T20:52:09

#V_(O_2)=181L#

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#