Nitrogen and hydrogen combine at high temperature,in presence of a catalyst,to produce ammonia.N2(g)+3H2(g)=2NH3.Assume 0.100 mol of N2 and 0.329 mol of H2 are present initially. After complete reaction,how many mols of ammonia are produced?How many moles of H2/N2 remain?
mol of N2 reqruired for 0.329 mol --> 1/3 * 0.329 = 0.109666 mol of N2, which we do not have, so N2 is limiting
mol of N2 reacted = 0.1
mol of N2 left = 0.1-0.1 = 0
mol of H2 reacted = 0.1*33 = 0.3 mol
mol of H2 left = 0.329 - 0.3 = 0.029 mol of H2
mol of NH3 produced = 0.1*2 = 0.2 mol of NH3
so..
total mol = 0.029+0.2 = 0.229 mol finally
Nitrogen and hydrogen combine at high temperature,in presence of a catalyst,to produce ammonia.N2(g)+3H2(g)=2NH3.Assume 0.100 mol of...
Nitrogen and hydrogen
combine at a high temperature, in the presence of a catalyst, to
produce ammonia.
N2(g)+3H2(g)⟶2NH3(g)N2(g)+3H2(g)⟶2NH3(g)
Assume 0.280 mol
N20.280 mol N2 and 0.880 mol H20.880 mol H2 are present
initially.
After complete reaction, how many moles of ammonia are
produced?
Nitrogen and hydrogen combine at a high temperature, in the presence of a catalyst, to produce ammonia. N2(g) + 3H2(g) + 2NH3(g) Assume 0.280 mol N, and 0.880 mol H, are present initially. After complete reaction, how...
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