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

The value is
`k =66\%`

Step-by-step 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 = (m_b)/(Z) * 100`

Here Z is the theoretical yield which is mathematically represented as

`Z = (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 =  (122.12 )/(92.14)  *  0.73`

=>
`Z = 0.968 \  g`

So

`k  =  (0.633)/(0.968)  * 100`

=>
`k  =66\%`