Question

0.73 grams of toluene was reacted with 2.0 grams of potassium permanganate in presence of 7.0 mL of 6 Molar potassium hydroxide and 30 mL of water. After refluxing for 1 hour the reaction mixture was treated with 6 Molar sulfuric acid to pH~ 2.0 followed by oxalic acid. On cooling this solution in an ice bath 0.633 grams of pure benzoic acid was obtained. Calculate the % yield of benzoic acid in this reaction.

Answer 1

Answer:

The value is $k =66\%$

Explanation:

From the question we are told that

The mass of toluene $m_t =0.73 \ g$

The mass of potassium permanganate is $m =2.0 \ g$

The volume of potassium hydroxide V = 7.0 mL

The concentration of potassium hydroxide C = 6 M

The mass of benzoic acid is $m_b = 0.633 \ g$

Generally the % yield of benzoic acid is mathematically represented as

$k = \frac{m_b}{Z} * 100$

Here Z is the theoretical yield which is mathematically represented as

$Z = \frac{E}{W} * m_t$

Here W is the molecular weight of product (benzoic acid) with value  

       W  = 92.14 \ g

E is the molecular weight of reactant (toluene)with a constant value  of  

    E =  122.12 g

So

     $Z = \frac{122.12 }{92.14} * 0.73$

=>    $Z = 0.968 \ g$

So

     $k = \frac{0.633}{0.968} * 100$

=>   $k =66\%$