A sample of gas contains 01500 mol of HCIg and 7500x102 mol

A sample of gas contains 0.1500 mol of HCI(g) and 7.500x102 mol of Br2(g) and occupies a volume of 12.4 L. The following reaction takes place: 2HCIg+ Br2(g) 2HBr(g)+ C2(g) Calculate the volume of the sample after the reaction takes place, assuming that the temperature and the pressure remain constant.

Solution

initial total moles of gas = 0.1500 + 0.07500

                            n1    = 0.225 mol

2 HCl   +   Br2    ---------> 2 HBr +   Cl2

2              1                        2             1

0.15        0.075

moles of HBr = 0.1500 mol

moles of Br2 = 0.15 / 2 = 0.075 mol

final moles = 0.15 + 0.075 = 0.225

here initial moles = final moles. so

volume is same

total volume of after reaction = 12.4 L