Potassium bromide reacts with chlorine to produce potassium chloride and bromine. How Many grams of bromine are produced when 15.0 g of potassium bromide are used?
Solution:
The balanced chemical equation:
2KBr + Cl2 → 2KCl + Br2
According to the equation above: n(Br2) = n(KBr)/2
Moles of KBr = Mass of KBr / Molar mass of KBr
The molar mass of KBr is 119.002 g mol-1.
Hence,
n(KBr) = (15.0 g) / (119.002 g mol-1) = 0.126 mol
n(Br2) = n(KBr)/2 = 0.126 mol / 2 = 0.063 mol
Moles of Br2 = Mass of Br2 / Molar mass of Br2
Mass of Br2 = Moles of Br2 × Molar mass of Br2
The molar mass of Br2 is 159.808 g mol-1.
Hence,
Mass of Br2 = (0.063 mol) × (159.808 g mol-1) = 10.0679 g = 10.07 g
Mass of Br2 = 10.07 g
OR (according to the equation):
(15.0 g KBr)×(1 mol KBr/119.002 g KBr)×(1 mol Br2/2 mol KBr)×(159.808 g Br2/1 mol Br2) = 10.07 g Br2
Answer: 10.07 grams of bromine (Br2) are produced.
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