posted by Drew .
When a sample of PCl5(g) (0.02087 mol/L) is placed in 83.00 L reaction vessel at 491.0 °C and allowed to come to equilibrium the mixture contains 103.0 grams of PCl3(g). What is the equilibrium concentration (mol/L) of Cl2(g)?
PCl5(g) = PCl3(g)+Cl2(g)
The answer is 9.037x10^-3 but I having trouble arriving to this answer.
Equilibrium concn PCl3 = 103.0 g.
M = (103.0/(137.33*83) = 0.009036 M
Since Cl2 and PCl3 are 1:1 in the equation, that must be the concn Cl2, also, at equilibrium.
Wot would happen if the pressure increase
In this reaction pcl5--> pcl3+ cl2