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.