A sample of gas contains 01400 mol of HCIg and 7000x102 mol

A sample of gas contains 0.1400 mol of HCI(g) and 7.000x102 mol of Br2(g) and occupies a volume of 13.9 L. The following reaction takes place: 2HCI(g) + Br2(g) 2HBr(g) + Cl2(g) Calculate the volume of the sample after the reaction takes place, assuming that the temperature and the pressure remain constant.

Solution

2HCl(g) + Br2(g) ------ 2HBr(g) + Cl2(g)

Number of moles of HCl = 0.140 moles

Number of moles of Br2 = 0.0700 moles

Since 2 moles of HCl will react with one mole of Br2, hence both of them will be fully used in the reaction, since 0.07 moles of Br2 will require 0.140 moles of HCl

Moles of HBr formed = Moles of HCl = 0.140 moles

Moles of Cl2 formed = Moles of Br2 = 0.070 moles

Using the ideal gas equation

PV = nRT (since temperature and pressure remain constant), so we can write

V1/n1 = V2/n2

V2 = V1 * n2/n1 = 13.9 * 0.21/0.21 = 13.9L

Hence the correct answer is 13.9L