Nitrogen and hydrogen combine at a high temperature, in the presence of a catalyst, to produce...
Nitrogen and hydrogen combine at a high temperature, in the presence of a catalyst, to produce ammonia. N2(g)+3H2(g)⟶2NH3(g) Assume 0.180 mol N2 and 0.564 mol H2 are present initially. How many moles of H2 remain?
Homework Answers
Balanced chemical equation is:
N2 + 3 H2 ---> 2 NH3 + CuS
1 mol of N2 reacts with 3 mol of H2
for 0.18 mol of N2, 0.54 mol of H2 is required
But we have 0.564 mol of H2
so, N2 is limiting reagent
According to balanced equation
mol of H2 reacted = (3/1)* moles of N2
= (3/1)*0.18
= 0.54 mol
mol of H2 remaining = mol initially present - mol reacted
mol of H2 remaining = 0.564 - 0.54
mol of H2 remaining = 2.4*10^-2 mol
Answer: 0.0240 mol
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Nitrogen and hydrogen combine at a high temperature, in the
presence of a catalyst, to produce...
Nitrogen and hydrogen combine at a high temperature, in the presence of a catalyst, to produce...
Nitrogen and hydrogen
combine at a high temperature, in the presence of a catalyst, to
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Assume 0.280 mol
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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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