Question

In a fixed volume container of volume 2 L, 13 g of Cl2 are mixed with 0.5 atm of HBr at 25 degrees Celsius. What is the final pressure of the bromine gas.

Cl2+2HBr->2HCl+Br2

Answer 1

For the reaction,

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

let us find the limiting reactant

moles of Cl2 present = 13 g/71 g/mol = 0.1831 mol

moles of HBr present (n) = PV/RT

P = 0.5 atm, V = 2 L, R = 0.0821 L.atm/K.mol, T = 25 oC + 273 = 298 K

So,

moles of HBr = 0.5 x 2/0.0821 x 298 = 0.0410 mol

Now,

moles of Br2 formed starting with Cl2 = 0.1831 mol

moles of Br2 formed starting with HBr = 0.0410/2 = 0.0205 mol

As moles of Br2 generated is less when started with HBr, HBr is the limiting reactant

final moles of Br2 formed = 0.0205 mol

final pressure of Br2 gas = nRT/V = 0.0205 x 0.0821 x 298/2 = 0.251 atm