Consider the following reaction: N2 + 3 H2 < - > 2 NH3. Initially, 4.25 moles of nitrogen gas and 6.33 moles of hydrogen gas are placed in a 3.35 L container. At equilibrium, 2.15 moles of NH3...

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Consider the following reaction: N2 + 3 H2 < - > 2 NH3. Initially, 4.25 moles of nitrogen gas and 6.33 moles of hydrogen gas are placed in a 3.35 L container. At equilibrium, 2.15 moles of NH3 was present. Calculate the number of moles of nitrogen and hydrogen at equilibrium.
Answered Same DayDec 20, 2021

Solution

Robert answered on Dec 20 2021
3 Votes
Solution:
N2 + 3H2 2 NH3
From the above relation it is known that 1 mole of N2 and 3 moles of H2 combines to give 2 moles of
NH3.
Let us suppose that, if 1 mole of N2 and 3 moles of H2 present in the reaction vessel. At equili
ium
the below relation satisfies
N2 + 3H2 ↔ 2 NH3
Before the reaction 1mole ...
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