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)

Molar Mass
PCl5(g) 208.24
PCl3(g) 137.33
Cl2(g) 70.906

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

To find the equilibrium concentration of Cl2(g), we need to use the stoichiometry of the balanced equation and the given information.

First, let's use the molar mass of PCl3(g) to find the number of moles of PCl3 present in the sample:
Molar mass of PCl3(g) = 137.33 g/mol
Mass of PCl3(g) = 103.0 g
Number of moles of PCl3(g) = Mass of PCl3(g) / Molar mass of PCl3(g)
= 103.0 g / 137.33 g/mol
= 0.750 mol

According to the balanced equation, for every mole of PCl3 reacted, one mole of Cl2 is produced. Therefore, the equilibrium concentration of Cl2(g) is equal to the number of moles of PCl3(g), which is 0.750 mol, since they have a 1:1 mole ratio.

Next, we need to calculate the volume of the reaction vessel in liters:
Volume of the reaction vessel = 83.00 L

Finally, we can calculate the equilibrium concentration of Cl2(g) using the formula:
Equilibrium concentration (mol/L) = Number of moles / Volume of the reaction vessel
= 0.750 mol / 83.00 L
= 0.00904 mol/L

Therefore, the equilibrium concentration of Cl2(g) is 0.00904 mol/L.