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 what is the formula for the limiting reagent? what amount of the excess reagent remains after the reaction is

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.

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