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.
