500 grams of a soil is taken and dried completely in bright sunshine. The mass of dried soil is found to be 450 grams. The perce
1 answer:
You might be interested in
Answer:
PCl₅ = 0.03 X 208 = 6.24g
PCl₃ = 0.05 X 137 =6.85 g
Cl₂ = 0.03X71 = 2.13 g
Explanation:
The equilibrium constant will remain the same irrespective of the amount of reactant taken.
Let us calculate the equilibrium constant of the reaction.
Kc=
Let us calculate the moles of each present at equilibrium
molar mass of PCl₅=208
molar mass of PCl₃=137
molar mass of Cl₂=71
moles of PCl₅ =
moles of PCl₃=
moles of Cl₂ =
the volume is 5 L
So concentration will be moles per unit volume
Putting values
Kc =
Now if the same moles are being transferred in another beaker of volume 2L then there will change in the concentration of each as follow
Initial 0.02 0.06 0.04
Change -x +x +x
Equilibrium 0.02-x 0.06+x 0.04+x
Conc. (0.02-x)/2 (0.06+x)/2 (0.04+x)/2
Putting values
0.024 =
Solving
x = -0.01
so the new moles of
PCl₅ = 0.02 + 0.01 =0.03
PCl₃ = 0.06-0.01 = 0.05
Cl₂ = 0.04-0.01 = 0.03
mass of each will be:
mass= moles X molar mass
PCl₅ = 0.03 X 208 = 6.24g
PCl₃ = 0.05 X 137 =6.85 g
Cl₂ = 0.03X71 = 2.13 g