HN3 decomposes by heat:
HN3 -- N2 + H2.
If 3.0 atm of pure HN3 is decomposed, what is the final total pressure and what are the partial pressures for nitrogen and hydrogen?

Question

HN3 decomposes by heat:
HN3 --> N2 + H2.
If 3.0 atm of pure HN3 is decomposed, what is the final total pressure and what are the partial pressures for nitrogen and hydrogen?

anonymous · Answer

Balance the equation to:

$$2HN _{3} \rightarrow 3N _{2} + H _{2}$$

anonymous · Answer

I know:
$$2 mol NH _{3} / 3 mol N _{2}$$
and
$$2 mol HN _{3}/1molH _{2}$$

anonymous · Answer

Oh, and assume the volume and temperature are constant.

anonymous · Answer

n HN3 = 3.0atm/0.08206 (l*atm/K*mol)?  Is that the next step?

anonymous · Answer

I've moved on from this one, but I plan on coming back to it so I can make sure that I have the concept.