How many moles of Cl2 are needed to produce 145.3 grams of Br2 from magnesium bromide?

To solve this problem, we first need to understand the balanced chemical equation given:

MgBr₂ (aq) + Cl₂ (g) → MgCl₂ (aq) + Br₂ (l)

From the equation, we see that 1 mole of Cl₂ produces 1 mole of Br₂. Therefore, the mole ratio of Cl₂ to Br₂ is 1:1.

Next, we need to convert grams of Br₂ into moles. The molar mass of Br₂ can be calculated as follows:

Since the atomic mass of bromine (Br) is approximately 79.9 g/mol, the molar mass of Br₂ will be:

Molar mass of Br₂ = 2 x 79.9 g/mol = 159.8 g/mol

Now, we will calculate the number of moles of Br₂ in 145.3 grams:

Moles of Br₂ = mass (g) / molar mass (g/mol)

Moles of Br₂ = 145.3 g / 159.8 g/mol ≈ 0.909 moles

Since the ratio of Cl₂ to Br₂ is 1:1, this means that the moles of Cl₂ needed would also be:

Moles of Cl₂ = 0.909 moles

Therefore, the chemist would need approximately 0.909 moles of Cl₂ to react with excess magnesium bromide to produce 145.3 grams of Br₂.