Respuesta :
Answer:
5.56 atm
Explanation:
N₂(g) + 3H₂(g) → 2NH₃(g)
10 moles of hydrogen gas was mixed with 5 moles of nitrogen gas.
we need to determine the limiting reagent
from the equation
1 mole of nitrogen requires 3 mole of hydrogen to produce 2 mole of ammonia
looking the reaction hydrogen is the limiting reagent because 5 moles of nitrogen will require 15 moles of hydrogen
1 mole of nitrogen ...... 3 moles of hydrogen ...... 2 moles of ammonia
x moles of nitrogen .......10 moles of hydrogen .... y moles of ammonia
10 / 3 moles of nitrogen is needed = 3.33 moles
10 × 2 / 3 moles of ammonia is produced = 6.667 moles
initial mole of nitrogen - consumed amount = residue
5 - 3.33 moles = 1.67 moles of nitrogen remained
total moles remaining = 1.67 moles of nitrogen + 6.667 mole of ammonia produced = 8.337 moles
now pressure is directly proportional to number of moles
15 moles ........... 10 atm
8.337 moles will yield = 8.337 moles × 10 atm / 15 moles = 5.56 atm
We have that the pressure (in atm) on the vessel [tex]p_2=7.78atm[/tex] be after the reaction takes place
From the Question we are told that
10 moles of hydrogen gas are mixed with 5 moles of nitrogen gas
Initial pressure exerted on the container is 10 atm.
Generally the Chemical Equation
[tex]N_2+3H_2 \rightarrow 2NH_3[/tex]
[tex]3H_2\ gives\ 2NH_3[/tex]
Therefore
[tex]5H_2=10/3NH_3[/tex]
Where
[tex]10-5/3=\frac{25}{3}[/tex]
Initial moles of gaseous mixture X
[tex]X=10+5\\\\X=15mol[/tex]
Finial moles of gaseous mixture X'
[tex]X'=\frfac{25}{3}+\frac{10}{3}\\\\X=\frac{35}{3}[/tex]
therefore at ideal gas state
[tex]\frac{p_1}{p_2}=\frac{n_1}{n_2}\\\\\frac{10}{p_2}=\frac{15}{\frac{35}{3}}[/tex]
[tex]p_2=7.78atm[/tex]
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