a.) Yes, your chemical equation representation is correct: Chlorine + potassium bromide → potassium chloride + bromine.
b) To solve this problem, we need to use the concept of stoichiometry and the principle of conservation of mass.
First, we need to determine the molar masses of the substances involved:
- Chlorine (Cl2) has a molar mass of 35.45 g/mol.
- Potassium bromide (KBr) has a molar mass of 119 g/mol.
- Potassium chloride (KCl) has a molar mass of 74.55 g/mol.
- Bromine (Br2) has a molar mass of 159.8 g/mol.
Next, we need to find the moles of chlorine, potassium bromide, and potassium chloride using their respective masses:
- Moles of chlorine = mass (g) / molar mass (g/mol) = 4.0 g / 35.45 g/mol = 0.113 mol
- Moles of potassium bromide = mass (g) / molar mass (g/mol) = 13.4 g / 119 g/mol = 0.112 mol
- Moles of potassium chloride = mass (g) / molar mass (g/mol) = 8.4 g / 74.55 g/mol = 0.113 mol
From the balanced chemical equation, we can see that the mole ratio between chlorine and potassium bromide is 1:1. Therefore, 0.113 mol of chlorine reacts with 0.113 mol of potassium bromide.
Since the reaction is assumed to go to completion, the limiting reactant is potassium bromide, and the number of moles of bromine produced is the same as the number of moles of potassium bromide: 0.112 mol.
Lastly, we can calculate the mass of bromine produced using its molar mass:
Mass of bromine = moles of bromine × molar mass of bromine
Mass of bromine = 0.112 mol × 159.8 g/mol = 17.9 g
Therefore, 17.9 grams of bromine are produced.