Answer:
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
Answer:
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
