Question

Cl_2 (g) +6F_2 (g) --> 2ClF_3 (g)

If a manufacturing plant needs 23.8 Lof ClF_3 at STP, what volume (in L) of chlorine gas is needed?

Answer 1

Cl2 (g) +6F2 (g) --> 2ClF3 (g)

No of mol of ClF3 = PV/RT

                    = (1*23.8/(0.0821*273.15))

                    = 1.06 mol

1 MOL Cl2 = 6 mol F2 = 2 mol ClF3

no of mol of Cl2 required = 1.06/2 = 0.53 mol

volume of Cl2 required = nRT/P

                         = 0.53*0.0821*273.15/1

                         = 11.9 L