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levacccp [35]
2 years ago
9

3 moles of nitrogen and 5 moles of hydrogen react to form ammonia. Use the balanced equation below to determine which is the lim

iting reactant.
N2 + 3H2 --> 2NH3
Chemistry
1 answer:
Nana76 [90]2 years ago
4 0

Answer:

H₂ is the limiting reactant

Explanation:

Balanced equations demonstrate the ratios of all reactants and products in terms of the number of atoms/molecules or moles;

In the given reaction:

1 mol of N₂ reacts with 3 mol of H₂ to produce 2 mol of NH₃

So the ratio of N₂ to H₂ to NH₃ will be 1:3:2

If we have 3 moles of N₂, we can apply the ratio to find that it will react with 9 moles of H₂ (3× the moles of N₂) to produce 6 moles of NH₃ (2× the moles of N₂);

Similarly, if we have 5 moles of H₂, then applying the ratio, we find that ⁵/₃ or 1.66... moles of N₂ (¹/₃× the moles of H₂) react with 5 moles of H₂ to produce ¹⁰/₃ or 3.33... moles of NH₃ (²/₃× the moles of H₂);

In order for all of the 3 moles of N₂ to react, it would require 6 moles of H₂;

There is only 5 moles of H₂ available, meaning there will be an excess of N₂;

5 moles of H₂ will react with 1.66... moles of the N₂, leaving 1.33... or ⁴/₃ moles of N₂ unreacted;

If the N₂ is in excess, then the H₂ is limited (i.e. the limiting reactant)

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Read 2 more answers
At 2000°C, the equilibrium constant for the reaction below is Kc = 4.10 ´ 10–4 . If 0.600 moles of NO is placed in a 1.0-L react
erastova [34]

Answer:

At equilibrium, the concentration of N_{2 (g)} is going to be 0.30M

Explanation:

We first need the reaction.

With the information given we can assume that is:

N_{2 (g)} + O_{2 (g)} ⇄ 2NO_{(g)}

If there is placed 0.600 moles of NO in a 1.0-L vessel, we have a initial concentration of 0.60 M NO; and no N_{2 (g)} nor  O_{2 (g)} present. Immediately, N_{2 (g)} andO_{2 (g)} are going to be produced until equilibrium is reached.

By the ICE (initial, change, equilibrium) analysis:

I: [N_{2 (g)}]=0   ;     [O_{2 (g)} ]= 0    ; [NO_{(g)}]=0.60M

C: [N_{2 (g)}]=+x   ;     [O_{2 (g)} ]= +x    ; [NO_{(g)}]=-2x

E: [N_{2 (g)}]=0+x   ;     [O_{2 (g)} ]= 0+x   ; [NO_{(g)}]=0.60-2x

Now we can use the constant information:

K_{c}=\frac{[products]^{stoichiometric coefficient} }{[reactants]^{stoichiometric coefficient} }

4.10* 10^{-4} =\frac{(0.60-2x)^{2}}{(x)*(x)}

4.10* 10^{-4}= \frac{(0.60-2x)^{2}}{x^{2} }

4.10* 10^{-4} * x^{2}= (0.60-2x)^{2}}

\sqrt{4.10* 10^{-4} * x^{2}}= \sqrt{(0.60-2x)^{2}}}

0.0202 x =0.60 - 2x

2x+0.0202x=0.60

x=\frac{0.60}{2.0202}= 0.30

At equilibrium, the concentration of N_{2 (g)} is going to be 0.30M

3 0
2 years ago
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