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WINSTONCH [101]
3 years ago
15

When 4.50 L of hydrogen gas react with an excess of nitrogen gas at standard temperature and pressure, how many liters of ammoni

a gas will be produced?
Chemistry
2 answers:
vichka [17]3 years ago
6 0
3H₂ + N₂ = 2NH₃

V(H₂)/3 = V(NH₃)/2

V(NH₃)=2V(H₂)/3

V(NH₃)=2*4.50/3=3.00 L

3.00 L
Delicious77 [7]3 years ago
6 0

<u>Answer:</u> The volume of ammonia gas produced will be 3 Liters.

<u>Explanation:</u>

<u>At STP:</u>

1 mole of a gas occupies 22.4 L of volume.

The chemical reaction for the formation of ammonia follows the equation:

N_2+3H_2\rightarrow 2NH_3

It is given that nitrogen gas is present in excess. Thus, hydrogen gas is considered as limiting reagent.

By Stoichiometry of the reaction:

3\times 22.4L of hydrogen produces 2\times 22.4L of ammonia gas.

So, 4.5 L of hydrogen gas will produce = \frac{2\times 22.4L}{3\times 22.4L}\times 4.5L=3L of ammonia gas.

Hence, the volume of ammonia gas produced will be 3 Liters.

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To calculate the new pressure, we can use Boyle’s law to relate these two scenarios (Boyle’s law is used because the temperature is assumed to remain constant). Boyle’s law is:

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