Nitrogen and hydrogen combine at a high temperature, in the presence of a catalyst, to produce ammonia. N 2 ( g ) + 3 H 2 ( g ) ⟶ 2 NH 3 ( g ) Assume 0.220 mol N 2 and 0.717 mol H 2 are present initially. After complete reaction, how many moles of ammonia are produced?

0.440 moles of NH₃ are produced

Explanation:

First of all, we need to determine the limiting reactant by the stoichiometry.

Equation reaction is: N₂(g) + 3H₂(g) ⟶ 2 NH₃(g)

1 mol of nitrogen needs 3 moles of hydrogen to react

Therefore 0.220 moles of N₂ will need (0.220 . 3) / 1 = 0.660 moles of H₂

As we have 0.717 moles of H₂ and we need 0.660, the hydrogen is the excess reagent, therefore, the N₂ is the limiting reactant

3 moles of H₂ need 1 mol of N₂ to react

Then, 0.717 moles of H₂ will react with (0.717 . 1) / 3 = 0.239 moles of N₂

We do not have enough N₂

After complete reaction → ratio is 1:2

1 mol of N₂ reacts to produce 2 moles of ammonia

Therefore 0.220 moles of N₂ will produce (0.220 . 2) / 1 = 0.440 moles of NH₃

distance, luminosity, brightness, radius, chemical composition, and temperature

consider the reaction below.

 no rx

at 500 k, the reaction is at equilibrium with the following concentrations.

[pci5]= 0.0095 m

[pci3] = 0.020

[ci2] = 0.020 m

what is the equilibrium constant for the given reaction?

0.042

0.42

2.4

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