Question

g Identify limiting reactants (mole ratio method). Close Problem Identify the limiting reactant in the reaction of bromine and chlorine to form BrCl, if 29.7 g of Br2 and 11.2 g of Cl2 are combined. Determine the amount (in grams) of excess reactant that remains after the reaction is complete.

Answer 1

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

                            $Moles of Br_{2} = \frac{weight}{Molecular weight} = \frac{29.7}{160} =0.18$

                            $Moles of Cl_{2} =\frac{11.2}{71} = 0.15 mole$

Using mole ratio method to find the limiting reactant and Excess reactant.

                            $\frac{Mole}{Stoichiometry} (for Br_{2} ) = \frac{0.18}{1} = 0.18$

                            $\frac{Mole}{Stoichiometry}(for Cl_{2}) =\frac{0.15}{1}=0.15$

So $Cl_{2}$ is a limiting reactant and Br₂ is excess reactant.

The amount of excess reactant = 0.18 - 0.15 = 0.03 mole = 4.8 g