(50 points) Reaction Stoichiometry, Limiting Reactants Ammonia (NH3) is produced
ID: 809685 • Letter: #
Question
(50 points) Reaction Stoichiometry, Limiting Reactants Ammonia (NH3) is produced by the reaction of hydrogen (H2) and nitrogen (N2) at about 800 K by the reaction 3 H2 + N2 2 NH3 Suppose a mixture of 20 kg of H2 and 100 kg of N2 are fed to a reactor. a. What is the limiting reactant? H2 b. What is the % excess reactant? 7.21% If 85 kg of NH3 are produced in the reaction, c. What is the conversion of the limiting reactant? 75% d. What is the conversion of the excess reactant? 70% e. Why is the conversion of the limiting reactant higher than the conversion of the excess reactant? You figure this out f. What is the composition of the product from the reactor (mole % H2, N2, and NH3)? 58.3% NH3, 29.2% H2, and 12.5% N2Explanation / Answer
b)see,you know that H2 is the limiting reactant here and N2 is the one in excess.No of moles of N2 are 100,000/28=3571.43and number of moles of H2=20,000/2=10,000.from the given equation one mole of N2 reacts with 3 moles of H2 which means the number of moles of N2 required is 1/3rd of the number of moles of H2 so the number of moles of N2 required =10,000/3=3,333.33 while no. of moles present=3571.43 so the excess of amount present=3571.43-3333.33=238.1.now percentage excess =(238.1/3,333.33)x100=7.21%
e)the conversion of the limiting reactant is higher because each mole of ammonia requires 3/2 moles of hydrogen but only 1/2 moles of nitrogen
f)No of moles of N2 are 100,000/28=3571.43and number of moles of H2=20,000/2=10,000 and no. of moles of NH3=85,000/17=5,000.as the NH3 produced is 5000 moles only so the H2 used=(3/2)x5000=7500 so remaining H2=2500 and remaining N2 =3571-(1/2)x5000-1071 so the required percentages are NH3=(5000/(5000+2500+1071))x100=(5000/8571)x100=58.3%,H2=2500/8571 x100=29.2% and N2=100-(58.3+29.2)=12.5%
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