Answer to Question #254633 in Chemistry for Kiman

Question

A 10.0 g-sample of potassium bromide is treated with perchloric acid solution. The reaction mixture is cooled and solid KClO4 is removed by filtering, then it is dried and weighed. KBr (aq)+HClO4 (aq) = KClO4 (s)+HBr (aq) The product weighed 8.8 g. What was the percent yield

Answer 1

The molar mass of KBr is 119.00 g/mol

The molar mass of KClO₄ is 138.55 g/mol

Solution:

Balanced chemical equation:

KBr(aq) + HClO₄(aq) → KClO₄(s) + HBr(aq)

According to stoichiometry:

(10.0 g KBr) × (1 mol KBr / 119 g KBr) × (1 mol KClO₄ / 1 mol KBr) × (138.55 g KClO₄ / 1 mol KClO₄) =

= 11.64 g KClO₄ - theoretical yield

11.64 g KClO₄ - theoretical yield

8.80 g KClO₄ - actual yield

%yield = (actual yield / theoretical yield) × 100%

Therefore,

%KClO₄ = (8.80 g / 11.64 g) × 100% = 75.6%

%KClO₄ = 75.6%

Answer: The percent yield is 75.6%