Calculate the volume of chlorine required to displace bromine from 500 g of 20%

Calculate the volume of chlorine required to displace bromine from 500 g of 20% potassium bromide solution. Subtract the mass of the salt formed.

1.Let’s find the mass of potassium bromide in solution by the formula:

w = m (substance): m (solution) × 100%,

hence m (substance) = (m (solution) × w): 100%.

m (KBr) = (500g × 20): 100% = 100g

or 500 g (solution) – 100%,

m g (solution) – 20%,

(500g × 20%): 100% = 100g.

2.Let’s find the amount of KBr.

n = m: M, where M is molar mass.

M (KBr) = 39 + 80 = 119 g / mol.

n = 100 g: 119 g / mol = 0.84 mol.

3. Let’s compose the equation of the reaction between potassium bromide and chlorine. Let’s find the quantitative ratios of potassium bromide and chlorine.

2 KBr + Cl2 = KCl + Br2.

2 moles of potassium bromide react with 1 mole of chlorine, that is, they are in a ratio of 2: 1, so the amount of chlorine will be 2 times less than the amount of potassium bromide.

n (Cl2) = 0.84: 2 = 0.42 mol.

Or:

2 mol KBr – 1 mol Cl2,

0.84 mol KBr – х mol Cl2,

x mol Cl2 = (0.84 mol × 1 mol): 2 mol = 0.42 mol.

4.Calculate the volume of chlorine by the formula:

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 = 22.4 L / mol × 0.42 mol = 8.96 L.

5. Find the mass of the formed salt – potassium chloride KCl.

2 mol KBr – 1 mol KCl,

0.84 mol KBr – х mol KCl,

x mol KCl = (0.84 mol × 1 mol): 2 mol = 0.42 mol.

Let’s calculate the mass of salt using the formula:

m = M × n, M (KCl) = 39 + 35 = 74 g / mol.

m = 74 g / mol × 0.42 mol = 31.08 g.

Answer: V (Cl2) = 8.96 L; m (KCl) = 31.08g.