What volume of "NH"_3NH3 at STP is produced if 25.0 "g"25.0g of "N"_2N2 is reacted with an excess of "H"_2H2 in the reaction "N"_2 + 3"H"_2 -> 2"NH"_3N2+3H2→2NH3?

1 Answer
Bio
Mar 16, 2016

40.0"L"40.0L

Explanation:

1 "mol"1mol of "N"_2N2 has mass of 28.0 "g"28.0g

Therefore, 25.0"g"25.0g of "N"_2N2 in moles is

frac{25.0"g"}{28.0 "g/mol"} = 0.893 "mol"25.0g28.0g/mol=0.893mol

Each molecule of "N"_2N2 will produce 2 molecules of "NH"_3NH3.

Therefore, the 2 times the amount of "NH"_3NH3 will be produced.

2 xx 0.893 "mol" = 1.79 "mol"2×0.893mol=1.79mol

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

Therefore, the volume occupied by 1.79 "mol"1.79mol of "NH"_3NH3 is

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

That's it. But bear in mind that you need to somehow remove the excess "H"_2H2 before the container will have a volume of 40.0 "L"40.0L.

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