til Tomorrow at 8 AM CST is allowed, with a one time 10% penalty to submitted sc

Question

til Tomorrow at 8 AM CST is allowed, with a one time 10% penalty to submitted score. Activity Information Use the References to access important values if needed for this question. For the following reaction, 57.0 grams of bromine are allowed to react with 29.5 grams of chlorine gas. bromine (g)+ chlorine (g) bromine monochloride (g) What is the maximum amount of bromine monochloride that can be formed? grams What is the FORMULA for the limiting reagent? What amount of the excess reagent remains after the reaction is complete? grams Submit Answer 3. Limiting Reagent: Smaller Amount Method: This is group attempt 3 of 5 Autosaved at 4:14 PM Back

Explanation / Answer

Balanced equation is

Br2 + Cl2 ---> 2 BrCl

number of moles of Br2 = 57.0 g / 159.808 g/mol = 0.357 mole

number of moles of Cl2 = 29.5 g / 70.906 g/mol = 0.416 mole

from the balanced equation we can say that

1 mole of Br2 requires 1 mole of Cl2 so

0.357 mole of Br2 will require 0.357 mole of Cl2

BUt we have 0.416 mole of Cl2 whihs is in excess and Cl2 is the excess reactant and

Br2 is the limiting reactant.

The formula of limiting reactant is Br2

from the balanced equation we can say that

1 mole of Br2 produces 2 mole of BrCl so

0.357 mole of Br2 will produce 0.714 mole of BrCl

mass of 1 mole of BrCl = 115.357 g

so the mass of 0.714 g of BrCl = 82.4 g

Therefore, the mass of BrCl produced would be 82.4 g

number of moles of excess reactant = 0.416 - 0.357 = 0.059 mole

mass of 1 mole of excess reactant = 70.906g so

the mass of 0.059 mole of excess reactant is 4.18 g

Therefore, the mass of excess reactant = 4.18 g

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