Calculate the mass of bromine formed by the interaction of 6.72 liters of chlorine with a 20%

| June 10, 2021 | education

Calculate the mass of bromine formed by the interaction of 6.72 liters of chlorine with a 20% potassium bromide solution, weighing 119 g.

The equation for the reaction of chlorine with potassium bromide:

2KBr + Cl2 = 2KCl + Br2

Amount of Cl2 substance:

v (Cl2) = V (Cl2) / Vm = 6.72 / 22.4 = 0.3 (mol).

KBr weight in solution:

m (KBr) = mr.-ra (KBr) * w (KBr) / 100 = 119 * 20/100 = 23.8 (g).

The amount of KBr substance:

v (KBr) = m (KBr) / M (KBr) = 23.8 / 119 = 0.2 (mol).

According to the reaction equation, 2 mol of KBr reacts with 1 mol of Cl2; therefore, only 0.2 / 2 = 0.1 mol of Cl2 should account for 0.2 mol of KBr. This means that KBr is taken in a deficiency, so the calculation is carried out on it.

The amount of bromine substance:

v (Br2) = v (KBr) / 2 = 0.2 / 2 = 0.1 (mol) and

mass of bromine formed:

m (Br2) = v (Br2) * M (Br2) = 0.1 * 160 = 16 (g).



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