Question

In the following reaction 1.55g of CI2 and 2.43g of KBr are mixed in solution. Determine the limiting and the excess reactants? How many grams of Br2 are formed? In lab 1.02g of Br2 was formed, what is the percent yield

Answer 1

2KBr + Cl₂ = 2KCl + Br₂

M (Cl₂) = 70.89 g/mol

M (Br₂) = 159.81 g/mol

M (KBr) = 119.00 g/mol

n = m/M

n (Cl₂) = 1.55/70.89 = 0.02 mol

n (KBr) = 2.43/119 = 0.02 mol

But, according to the equation, n (KBr) = 2 x n (Cl₂).

Therefore, KBr is the limiting reaction in this case. Cl₂ is the excess reactant.

m (Br₂) = n (Br₂) x M (Br₂)

n (Br2) = n (KBr) / 2 = 0.02 / 2 = 0.01 mol

m (Br₂) = 0.01 x 159.81 = 1.60 g

% (Br₂) = 1.02/1.60 x 100 = 63.75%