What mass of bromine will be released when passing excess chlorine through a solution of potassium

What mass of bromine will be released when passing excess chlorine through a solution of potassium bromide with a mass of 25 g. and a concentration of 7%?

When excess chlorine interacts with a potassium bromide solution, the reaction products are potassium chloride and free bromine. First, let’s find how much pure 100% potassium chloride is in the solution. To do this, we multiply the mass content of KCl by the mass of the solution:
m = 25 * 0.07 = 1.75 grams.
Next, we write the equation of the interaction reaction:
2KBr + Cl2 = 2KCl + Br2
To find the mass of the resulting bromine, let’s make the proportion:
1.75 – x
2 * 119 – 160
Whence the mass of bromine is equal to:
x = 1.75 * 160/2 * 119 = 1.176 grams