Answer:
(i) Cl₂ is a limiting reactant
(ii) The amount of excess reactant = 4.8 g
Explanation:
Br₂(g) + Cl₂(g) → 2 BrCl(g) ---------------------------(i)
Calculation of no. of moles
[tex]Moles of Br_{2} = \frac{weight}{Molecular weight} = \frac{29.7}{160} =0.18[/tex]
[tex]Moles of Cl_{2} =\frac{11.2}{71} = 0.15 mole[/tex]
Using mole ratio method to find the limiting reactant and Excess reactant.
[tex]\frac{Mole}{Stoichiometry} (for Br_{2} ) = \frac{0.18}{1} = 0.18[/tex]
[tex]\frac{Mole}{Stoichiometry}(for Cl_{2}) =\frac{0.15}{1}=0.15[/tex]
So [tex]Cl_{2}[/tex] is a limiting reactant and Br₂ is excess reactant.
The amount of excess reactant = 0.18 - 0.15 = 0.03 mole = 4.8 g