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# Scan_Doc0018 - 3.n The Concept of Limiting Reagent 111...

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Always determine which reactant is limiting. 3.n The Concept of Limiting Reagent 111 Thus 8.93 X 10 2 mol N2requires 2.68 X 10 3 mol H2to react completely. However, in this case, only 2.48 X 10 3 mol H2is present. This means that the hydrogen will be consumed before the nitrogen. Thus hydrogen is the limiting reactant in this particular situation, and we must use the amount of hydrogen to compute the quantity of ammonia formed: 2 mol NH3 2.48 X 10 3 mel-tf;: X = 1.65 X 10 3 mol NH3 3mel-tf;: Converting moles to kilograms gives 1.65 X 10 3 ~ X 17.0 g NH3 = 2.80 X 10 4 g NH3 = 28.0 kg NH3 l~ Note that to determine the limiting reactant, we could have started instead with the given amount of hydrogen and calculated the moles of nitrogen required: 1 mol N2 2.48 X 10 3 mel-tf;: X = 8.27 X 10 2 mol N2 3mel-tf;: Thus 2.48 X 10 3 mol H2requires 8.27 X 10 2 mol N 2 .Since 8.93 X 10 2 mol N2is actually present, the nitrogen is in excess. The hydrogen will run out first, and thus again we find that hydrogen limits the amount of ammonia formed.

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