Question

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

Answer 1

Answer:

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

Answer 2

Answer:

15.81 g

Explanation:

Step 1: Write the balanced equation

F₂ + 2 KBr → 2 KF + Br₂

Step 2: Calculate the moles corresponding to 23.55 g of potassium bromide

The molar mass of potassium bromide is 119.00 g/mol.

$23.55g \times \frac{1mol}{119.00g} =0.1979mol$

Step 3: Calculate the moles of bromine produced

The molar ratio of KBr to Br₂ is 2:1.

$0.1979molKBr \times \frac{1molBr_2}{2molKBr} =0.09895molBr_2$

Step 4: Calculate the mass corresponding to 0.09895 moles of bromine

The molar mass of bromine is 159.81 g/mol.

$0.09895 mol \times \frac{159.81g}{mol} =15.81 g$