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

(i) Cl₂ is a limiting reactant

(ii) The amount of excess reactant = 4.8 g

Step-by-step explanation:

Br₂(g) + Cl₂(g) → 2 BrCl(g) ---------------------------(i)

Calculation of no. of moles

`Moles of Br_(2) = (weight)/(Molecular weight) = (29.7)/(160) =0.18`

`Moles of Cl_(2) =(11.2)/(71) = 0.15 mole`

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

`(Mole)/(Stoichiometry) (for Br_(2) ) = (0.18)/(1) = 0.18`

`(Mole)/(Stoichiometry)(for Cl_(2)) =(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