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.

Limiting reactant: Cl₂

4.47 g of Br₂ will remain after the reaction is complete.

Explanation:

Our reactants are: Br₂ and Cl₂

The product is the BrCl

Therefore the reaction is: Br₂ + Cl₂ → 2BrCl

We need the moles of each reactant, let's convert the mass to moles  

29.7 g / 159.80 g/mol = 0.186 moles

11.2 g / 70.9 g/mol = 0.158 moles

Ratio is 1:1, so we need the same amount of each reactant. As we need 0.186 moles of chlorine ( beacuse it is the amount of bromine), and we only have 0.158 moles, Cl₂ is the limiting reactant.

So the excess reactant is definately the Br₂. We need 0.158 moles, and we have 0.186 moles, so (0.186 - 0.158) = 0.028 moles will remain after the reaction is complete.

Let's convert the moles to mass: 0.028 mol . 159.8 g /1mol = 4.47 g of Br₂

There will remain 4.47 grams of Br2

Explanation:

Step 1: data given

Mass of Br2 = 29.7 grams

Molar mass of Br2 = 159.81 g/mol

Mass of Cl2 = 11.2 grams

Molar mass of Cl2 = 70.9 g/mol

Step 2: The balanced equation

Br2 + Cl2 → 2BrCl

Step 3: Calculate moles Br2

Moles Br2 = 29.7 grams / 159.81 g/mol

Moles Br2 = 0.186 moles

Step 4: Calculate moles Cl2

Moles Cl2 = 11.2 grams / 70.9 g/mol

Moles Cl2 = 0.158 moles

Step 5: Calculate limiting reactant

For 1 mol Br2 we need 1 mol CL2 to produce 2 moles BrCl

Cl2 is the limiting reactant. It will completely be consumed (0.158 moles)

Br2 is in excess. There will react 0.158 moles. There will remain 0.186 - 0.158 = 0.028 moles Br2

Step 6: Calculate moles of Br2 remaining

Moles Br2 = 0.028 moles * 159.81 g/mol

Moles Br2 = 4.47 grams

There will remain 4.47 grams of Br2

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