[K_(c)" for "PCl_(5)rightleftharpoons PCl_(3)+Cl_(2)" is "0.04" at "25^(@)C" .How many moles "],[" of "PCl_(5)" must be added to a "3" litre flask to obtain a "Cl_(2)" of "],[" concentration "0.15M" ."]

At 250^@C , K_c for PCl_5(g)hArr PCl_3(g)+Cl_2(g) is 0.04 . How many moles of PCl_5 must be added to a 3-L flask to obtain 0.15M Cl_2 at equilibrium?

The equilibrium constant for the equilibrium PCl_(5)(g) hArr PCl_(3)(g) + Cl_(2)(g) at a particular temperature is 2 xx 10^(-2)" mol dm"^(-3) . The number of moles of PCI_(5) that must be taken in a one-litre flask at the same temperature to obtain a concentration of 0.20 mol of chlorine at equilibrium is

For the reaction PCl_(5)hArr PCl_(3)+Cl_(2) in gaseous phase, K_(c)=4 . In a 2L flask, there are 2 moles each of PCl_(3) and Cl_(2) and 0.5 mole of PCl_(5) . Equilibrium concentration of PCl_(5) is

PCl_(5)(g)hArr PCl_(3)(g)+Cl_(2)(g) , In above reaction, at equilibrium condition mole fraction of PCl_(5) is 0.4 and mole fraction of Cl_(2) is 0.3. Then find out mole fraction of PCl_(3)