What volume of #"NH"_3# at STP is produced if #25.0 "g"# of #"N"_2# is reacted with an excess of #"H"_2# in the reaction #"N"_2 + 3"H"_2 -> 2"NH"_3#?

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Answer 1 · 2016-03-16T03:45:38

#40.0"L"#

#1 "mol"# of #"N"_2# has mass of #28.0 "g"#

Therefore, #25.0"g"# of #"N"_2# in moles is

#frac{25.0"g"}{28.0 "g/mol"} = 0.893 "mol"#

Each molecule of #"N"_2# will produce 2 molecules of #"NH"_3#.

Therefore, the 2 times the amount of #"NH"_3# will be produced.

#2 xx 0.893 "mol" = 1.79 "mol"#

One mole of an ideal gas has a volume of #22.4"L"# at STP.

Therefore, the volume occupied by #1.79 "mol"# of #"NH"_3# is

#1.79 "mol" xx 22.4 "L/mol" = 40.0 "L"#

That's it. But bear in mind that you need to somehow remove the excess #"H"_2# before the container will have a volume of #40.0 "L"#.