Question

How can you manufacture
(i) $KMn{{O}_{4}}$ from $Mn{{O}_{2}}$
(ii) ${{K}_{2}}C{{r}_{2}}{{O}_{7}}$ from chromite ore

Answer 1

Chromite ore is going to contains chromium in the form of $N{{a}_{2}}Cr{{O}_{4}}$ and the manganese oxide ($Mn{{O}_{2}}$ ) is called as pyrolusite. There is a requirement of the chemicals to treat the ores to convert into a desired form.

Complete answer:
- In the question it is asked to prepare$KMn{{O}_{4}}$ from $Mn{{O}_{2}}$ and ${{K}_{2}}C{{r}_{2}}{{O}_{7}}$ from chromite ore.
(i) First, we discuss the preparation of the$KMn{{O}_{4}}$ from $Mn{{O}_{2}}$ and it is as follows.
- It is a two-step process to prepare $KMn{{O}_{4}}$ from $Mn{{O}_{2}}$
Step-1: $2Mn{{O}_{2}}+4KOH+{{O}_{2}}\xrightarrow{\Delta }2{{K}_{2}}Mn{{O}_{4}}+2{{H}_{2}}O$
- In the step-1 one mole of manganese oxide is going to react with 4 moles of potassium hydroxide and one mole oxygen at higher temperature forms potassium manganate as the product.
Step-2: $2{{K}_{2}}Mn{{O}_{4}}+C{{l}_{2}}\to KMn{{O}_{4}}+2KCl$
- In step-2 the formed potassium manganate reacts with chlorine and forms potassium permanganate as the product.
(ii) Preparation of ${{K}_{2}}C{{r}_{2}}{{O}_{7}}$ from chromite ore.
- It is a two-step process to prepare ${{K}_{2}}C{{r}_{2}}{{O}_{7}}$ from chromite ore.
Step-1: $2N{{a}_{2}}Cr{{O}_{4}}+{{H}_{2}}S{{O}_{4}}\to N{{a}_{2}}C{{r}_{2}}{{O}_{7}}+N{{a}_{2}}S{{O}_{4}}+{{H}_{2}}O$
- In the first step sodium chromate is going to be treated with sulphuric acid to get sodium dichromate.
Step-2: $N{{a}_{2}}C{{r}_{2}}{{O}_{7}}+2KCl\to {{K}_{2}}C{{r}_{2}}{{O}_{7}}+2NaCl$
- In the step-2 sodium dichromate is going to react with potassium chloride and forms potassium chromate as a final product.

Note:
The molecular formulas of the respective chemical ores must be known to prepare the respective compounds. Without knowing the molecular formulas of the chemicals which are present in the ores we can not write the preparation of various chemicals from the respective ores.