Reply
Fri 7 Feb, 2014 04:11 pm
Nitrogen and hydrogen gases react to form ammonia gas as follows:
N2(g)+3H2(g)→2NH3(g){\rm{N}}_2 (g) + 3{\rm{H}}_2 (g) \;\rightarrow \;2{\rm{NH}}_3 (g)
At a certain temperature and pressure, 1.4LL of N2{\rm{N}}_2 reacts with 4.2LL of H2{\rm{H}}_2.
If all the N2{\rm{N}}_2 and H2{\rm{H}}_2 are consumed, what volume of NH3{\rm{NH}}_3, at the same temperature and pressure, will be produced?
@bbartocha,
Enough volume to go fffffffffttt for just over ten seconds.
@bbartocha,
At the same pressure and temperature the the volume of a gas is proportional to its molar quantity.
Ideal Gas Law
PV=nRT so
n=PV/RT=(P/RT)V
So if You had 1.4 L of Nitrogen gas and 4.2L (3*1.4) of Hydrogen gas
The by the Reaction
N2(g)+3H2(g)-> 2NH3(g)
So you would produce twice as much gaseous ammonia (by volume) as nitrogen----
or 2.8L of NH3(g)
Rap
Stumble It!
Tweet This
Bookmark on Delicious
Share on Facebook
Share on MySpace