The Grass Doctor

Can you help set up this gas problem please?

Can someone help me with the problem below please? I'm in an introductory chem class and have no idea how to set it up. We just learned some gas laws, so all I know is PV = NRT. I don't know how to in other variables (especially the minutes). Could someone explain please as detailed as possible. This is the problem: Urea (H2NCONH2) is used extensively as a nitrogen source in fertilizers. It is produced commercially from the reaction of ammonia and carbon dioxide. 2 NH3(g) + CO2(g) --> (heat and pressure) --> H2NCONH2(s) + H2O(g) Ammonia gas at 223°C and 90. atm flows into a reactor at a rate of 450. L/min. Carbon dioxide at 223°C and 41 atm flows into the reactor at a rate of 600. L/min. What mass of urea is produced per minute by this reaction assuming 100% yield?

Public Comments

1. sorry i would help if i had more time but i gotta go. i should be able to do this easily too becuase i just did this chapter in chem are you in high school or college? just curious to see if i will have this again in college. im guessing it will have something to do with effusion equation (rate1)/(rate 2) = square root ((molar mass 1)/(molar mass 2)) R=.08206 R is just a standard value P=pressure (atm) T=temp in kelvins K=C(celcius) +273 hope this helps a litte
2. The minutes are not relevant in this case as everything is per minute. The problem can be rewritten as: Assuming 100% conversion how much urea is produced from 450 litres ammonia at 223°C and 90 atm and 600 litres carbon dioxide at 223°C and 41 atm. What has to be done is to find the moles of ammonia and carbon dioxide the above translates to. It is then a trivial matter to find the moles urea produced. Multiply by the molwt of urea to get the mass of urea produced. Ammonia at STP= 450 x (273/496) x 90 = 22291 litres Moles ammonia= 22291/22.4 =995.15 gm moles carbon dioxide at STP= 600 x (273/496) x 41= 13540 litres Moles carbon dioxide = 13450/22.4 = 604.5 gm moles Since 995.15 gm moles ammonia require only 497.575 moles of carbon dioxide, there is excess carbon dioxide. At 100% yield urea produced = 497.575 gm moles. This becomes 497.575 x 78 /1000 = 38.81 kg/min of urea.