Respuesta :
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
