Assuming 100% yield, calculate the maximum volume of ammonia that could be produced from 1200dm3 of hydrogen, measured at room temp and pressure.

Okay so first always write out the equation...N₂(g) + 3H₂(g) ⇌ 2NH₃(g)Now we need to ask ourselves what equations we know are relevant to the question. The question gives us a volume and states the conditions as room temperature and pressure (RTP). Therefore we need to use the fact that 1 mole of gas occupies 24dm³ at RTP:1200/24 = 50 moles of hydrogen gasNext we use the ratio in the equation to calculate the theoretical number of moles of ammonia. For every three moles of hydrogen we get 2 moles of ammonia so:(50/3)x2= 33.33 moles of ammonia (Note: if we didn't have 100% yield we would take this into account at this stage in the calculation)Then convert this back into volume using the same equation we used earlier:33.33 x 24 = 800 dm³ of ammoniaFor this question we could have also used an intuitive approach based upon the fact that the conditions do not change. From this you can deduce that there will the 2/3 of the volume of ammonia as hydrogen. However always using the longer method reduces the potential for any possible confusion.