Question

For the following reaction, 33.7 grams of bromine are allowed to react with 13.0 grams of chlorine gas.

Answer 1

Explanation:

The reaction is given as;

Br2(g) + Cl2(g) ----> 2BrCl(g)

From the equation;

1 mol of Br2 reacts with 1 mol of Cl2

Converting the masses given to moles, using the formular;

Number of moles = Mass / Molar mass

Br2;

Number of moles = 33.7 g / 159.808 g/mol = 0.21088 mol

Cl2;

Number of moles = 13.0 g / 70.906 g/mol = 0.18334 mol

From the values;

0.18334 mol of Cl2 would react with 0.18334 mol of Br2 with an excess of 0.02754 mol of Br2

What is the maximum amount of bromine monochloride that can be formed? __________grams

1 mol of Cl2 produces 2 mol of Bromine Monochloride

0.18334 mol of Cl2 would produce x

Solving for x;

x = 0.18334 * 2 = 0.36668 mol

Converting to mass;

Mass = Number of moles * Molar mass = 0.36668 mol * 115.357 g/mol

Mass = 42.299 g

What is the FORMULA for the limiting reagent?

The limiting reagent is Cl2 as it determines the amount of product formed. The moment the reaction uses up Cl2, the reaction stops.

What amount of the excess reagent remains after the reaction is complete? __________grams

The excess reagent is Br2

The number of moles left is;

0.02754 mol of Br2

Converting to mass;

Mass = Number of moles * Molar mass = 0.02754 mol  * 159.808 g/mol

Mass = 4.401 g