An excess of hydrogen reacts with 14.0 g of #N_2#. What volume of ammonia will be produced at STP?

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Answer 1 · 2016-05-03T13:19:44

#V=33.6L#

The reaction between hydrogen and nitrogen is the following:

#3H_2(g)+N_2(g)->2NH_3(g)#

Using the mass of nitrogen #m=14.0g#, we can first find the number of mole of ammonia that will be produced:

#?molNH_3=14.0cancel(gN_2)xx(1cancel(molN_2))/(28.0cancel(gN_2))xx(3molNH_3)/(1cancel(molN_2))=1.50molNH_3#

It is known that at STP, one mole of gas occupies a volume of #22.4L#. This is called the molar volume.

Thus, the ammonia produced will occupy the following volume:

#?L=1.50cancel(molNH_3)xx(22.4L)/(1cancel(molNH_3))=33.6 L#