How do you calculate this percentage yield?

Under aerobic conditions, 200kg of hydrogen reacts with excess nitrogen in the Haber process to produce 650kg of ammonia.

calculate the percentage yield?



answer should come to 57.37%



please show all working



%26amp; also how do you do this question: 0.6 g of I2 dissolved in 10cm3 of KI (aq) and 10cm3 of CHCl3(l). Calculate the concentration of iodine in g l-1 continued in CHCl3.



thanksHow do you calculate this percentage yield?
Percentage yield in general terms is:



% yield = actual amount produced / theoretical yield * 100%



The actual amount produced was 650 kg. So we need to find the theoretical yield.



The chemical equation for this reaction would be:



3H2 + N2 = 2NH3



The extent of this reaction is dictated by the amount of H2 gas we have (as N2 is present in excess), therefore if we have 200 kg of H2 gas we have X moles of H2 gas. Finding this X is below:



200 kg = 200,000 grams, 200,000 grams / roughly 2 g/mole = 100,000 moles of H2 gas.



Because the stoichiometric coefficients between H2 and NH3 are 3:2 we have to multiply this number of moles of H2 by 2/3 to get the number of moles of NH3 theoretically produced from this amount of hydrogen gas.



Therefore theoretically we should produce 66,667 moles of NH3 gas.



66,667 moles at 17.031 grams/mole will give us 1,135,405.677 grams of NH3 or 1135.4 kg of Ammonia which should be theoretically produced.



Therefore:



Actual Yield / Theoretical Yield * 100% = PY = 650 kg / 1,135.4 kg * 100% = 57.25%



That is how you are supposed to do it. The small variation between my number and the number you quoted could be because I used a few rough estimates, but that should be the way to do it.



The last part I don't quite understand what you are asking, sorry.



Hope this helps!