How many grams of bromine are produced when 23.55 g of potassium bromide are consumed?
F2 + 2KBr → 2KF + Br2

15.8 g

Explanation:

In every stoichiometry problem, we must first note the equation of the reaction as it serves as a guide in solving the problem.

F2 + 2KBr → 2KF + Br2

No of moles of potassium bromide corresponding to 23.55g = mass/ molar mass

Molar mass of potassium bromide= 119.002 g/mol

No of moles = 23.55g / 119.002 g/mol

No of moles = 0.1979 moles

From the reaction equation;

If 2 moles of potassium bromide yields 1 mole of bromine

0.1979 moles of potassium bromide will yield 0.1979 moles × 1 mole/ 2 moles= 0.09895 moles of bromine

Mass of bromine produced= number of moles × molar mass

Molar mass of bromine= 2(79.904)gmol-1

Mass of bromine= 0.09895 moles × 2(79.904) gmol-1

Mass of bromine= 15.8 g