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theoretical yeild question?

Need theorteical yeild of this. I think I balanced it correctly already. Reaction for the production of urea, a fertilizer. CO2 + 2NH3 ---> NH2CONH2 + H2O Determine the teoretical yield in kg og urea if 2.65 kg of CO2 and 1.25 kg of NH3 are reacted? Please help, this is for final review and I am stuck!

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  1. wow have u paid any attention in your chem class at all? the theorteical yeild of water is 0.661 Kg and the theorteical yeild of NH2CONH2 is 2.203 Kg
  2. 2.65 kg CO2 = 2,650 g CO2 1.25 kg NH3 = 1,250 g NH3 Convert to moles. The molar mass of CO2 is 44.0 g/mol. 2,650 g CO2 x (1 mol / 44.0 g) = 60.2 mol CO2 The molar mass of NH3 is 17.0 g/mol. 1,250 g NH3 x (1 mol / 17.0 g ) = 73.5 mol NH3 Find the limiting reagent. Assume all the CO2 is used up, find the amount of NH3 needed. 2 mol NH3 reacts with 1 mol CO2. 60.2 mol CO2 x (2 mol NH3 / 1 mol CO2) = 120.4 mol NH3 Since there are only 73.5 mol of NH3, it must be the limiting reagent. Suppose you started with the NH3. 73.5 mol NH3 x (1 mol CO2 / 2 mol NH3 ) = 36.8 mol CO2 needed. Since there are 60.2 mol CO2, there is an excess amount of CO2 to use up all the NH3. Again, NH3 must be the limiting reagent. Use the limiting reagent to find the theoretical yield. 2 mol NH3 produces 1 mol urea. 73.5 mol NH3 x (1 mol urea / 2 mol NH3) = 36.8 mol urea The molar mass of urea is 60.0 g/mol 36.8 mol x (60.0 g / 1 mol) = 2210 g urea = 2.21 kg urea.
  3. Yep, you balanced the equation right. The next step would be to find what is the limiting reactant (the reactant that is used up completely). Use Dimensional analysis 2650gCO2 x 1mol CO2/44g x 1mol Urea/1mol CO2 = 60.2 mol Urea 1250g NH3 x 1mol NH3/ 17g x 1mol Urea/2mol =36.8 mol Urea. Now you can see that NH3 is the limiting reactant because it produces the least amount of Urea. Now the theoretical yield is going to be the amount of Urea that can be made if the reaction goes to completion with 100% success/yield. We already found the molar amount that could be theoretically produced and that is 36.8 mol. Now all we need to do is convert 36.8 mol Urea to kg 36.8 mol Urea x 60g/1mol Urea =2208g = 2.21kg
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