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
<em>What is the maximum amount of bromine monochloride that can be formed? __________grams</em>
<em />
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
<em />
<em>What is the FORMULA for the limiting reagent?</em>
<em />
The limiting reagent is Cl2 as it determines the amount of product formed. The moment the reaction uses up Cl2, the reaction stops.
<em>What amount of the excess reagent remains after the reaction is complete? __________grams</em>
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
