posted by Ezra on .
Nitrogen & hydrogen goes react to form ammonia gas as follows:
At a certain temperature & pressure, 1.2L of N2 reacts with 3.6L of H2. If all the N2 and H2 are consumed, what volume of NH3, at the same temperature & pressure will be produced?
I tried working it out with PV=nRT to find the moles of each reactant and then combined them to use PV=nRT again to find the volume of ammonia, but I'm not sure if that's the correct answer or procedure.
An explanation would be great as to how I'm supposed to go about this problem and why. Thanks
There is a long way and a shortcut way. First the long way which will work on all problems.
This is a simple (as opposed to limiting reagent stoichiometry problem) stoichiometry problem. (USUALLY we know it is a limiting reagent problem when amounts for BOTH reactants are given; however, in this one the problem states that BOTH reactants are consumed entirely.)
N2 + 3H2 ==> 2NH3
First you do PV = nRT to obtain mols N2 and H2. You started off on the right foot although you needed only one of them.
Next, using the coefficients in the balanced equation, use either mols N2 or mols H2 to convert to mols NH3.
Then you can use PV = nRT and convert mols NH3 to volume.
The short way. When all are gases involved we need not convert to mols; i.e., we can use volume directly.
Therefore, convert 1.2L N2 to L NH3.
1.2L N2 x (2 mols NH3/1 mol N2) = 2.4 L NH3.
Or you could use L H2 the same way.
3.6 L H2 x (2 mols NH3/3 mol H2) = 2.4 L NH3.
Thanks so much, that really helped me understand it.