Question

For the following reaction, 60.3 grams of bromine are allowed to react with 25.0 grams of chlorine gas.

bromine (g) + chlorine (g) bromine monochloride (g)

What is the maximum amount of bromine monochloride that can be formed?____g

What is the FORMULA for the limiting reagent?


What amount of the excess reagent remains after the reaction is complete?____g

Answer 1

Balanced chemical reaction: Br₂(g) + Cl₂(g) → 2BrCl(g).

1) Answer is: the maximum amount of bromine monochloride is 81.21 grams.

m(Br₂) = 60.3 g; mass of bromine.

n(Br₂) = m(Br₂) ÷ M(Br₂).

n(Br₂) = 60.3 g ÷ 159.8 g/mol.

n(Br₂) = 0.377 mol; amount of bromine.

m(Cl₂) = 25.0 g; mass of chlorine gas.

n(Cl₂) = m(Cl₂) ÷ M(Cl₂).

n(Cl₂) = 25 g ÷ 70.9 g/mol.

n(Cl₂) = 0.352 mol; amount of chlorine gas.

2) Formula for the limiting reagent is Cl₂, because all chlorine will react.

From balanced chemical reaction: n(Br₂) : n(BrCl) = 1 : 2.

n(BrCl) = 2 · 0.352 mol.

n(BrCl) = 0.704 mol; amount of bromine monochloride.

m(BrCl) = n(BrCl) · M(BrCl).

m(BrCl) = 0.704 mol · 115.357 g/mol.

m(BrCl) = 81.21 g; mass of bromine monochloride.

3) From balanced chemical reaction: n(Cl₂) : n(Br₂) = 1 : 1.

n(Br₂) = n(Cl₂) = 0.352 mol; amount of substance that will react.

Δn(Br₂) = 0.377 mol - 0.352 mol.

Δn(Br₂) = 0.025 mol.

Δm(Br₂) = 0.025 mol · 159.8 g/mol.

Δm(Br₂) = 3.995 g; amount of the excess reagent remains after the reaction is complete.