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 $"N"_2 + 3"H"_2 -> 2"NH"_3$ ?

Question

15864 views around the world

You can reuse this answer
Creative Commons License

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 $40.0 "L"$ .

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