Nitrogen and hydrogen combine at high temperature, in the presence of a catalyst

Question

Nitrogen and hydrogen combine at high temperature, in the presence of a catalyst, to produce ammonia. N2(g) + 3H2(g) --> 2NH3(g) Assume 0.260 mol of N2 and 0.821 mol of H2 are present Initially. After complete reaction, how many moles of ammonia are produced? How many moles of H2 remain? How many Moles of N2 remain? What is the limiting reactant?

Explanation / Answer

N2 + 3H2 -----> 2NH3 1 mole N2 reacts with 3 mole H2 0.26 mol N2 requires 0.26 x3 =0.78 mol H2 1 mole N2 gives 2 mol NH3 0.26 mol N2 gives 2x0.26 =0.52 mol NH3 Ammonia formed moles = 0.52 moles of H2 remaining = 0.821-0.78 =0.041 moles of N2 remaining = 0 N2 is limiting reactant. Since N2 is present in less moles in reaction when compared to other reactant H2 which is excess.

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