During the electrolysis of an aqueous solution of potassium bromide, bromine with a mass of 8 g

| August 8, 2021 | education

During the electrolysis of an aqueous solution of potassium bromide, bromine with a mass of 8 g was released at the anode.

Let’s find the amount of bromine substance by the formula:

n = m: M.

M (Br2) = 80 × 2 = 160 g / mol.

n = 8 g: 160 g / mol = 0.05 mol.

Let’s compose the reaction equation, find the quantitative ratios of substances.

2 KBr + 2H2O = H2 + 2KOH + Br2.

According to the reaction equation, there is 1 mole of hydrogen per mole of bromine. Substances are in quantitative ratios of 1: 1.

The amount of substance will be the same.

n (H2) = n (Br2) = 0.05 mol.

Let’s find the volume of hydrogen.

V = Vn n, where Vn is the molar volume of gas, equal to 22.4 l / mol, and n is the amount of substance.

V = 0.05 mol × 22.4 L / mol = 1.12 L.

Answer: V = 1.12 liters.



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