Question

What will be the product when $ {{K}_{2}}C{{r}_{2}}{{O}_{7}} $ reacts with cone. $ {{H}_{2}}S{{O}_{4}}, $ i.e. $ {{K}_{2}}C{{r}_{2}}{{O}_{7}}+2{{H}_{2}}S{{O}_{4}}\xrightarrow{Cold} $

• A. $ C{{r}_{2}}{{O}_{3}} $
• B. $ Cr{{O}_{3}} $
• C. $ C{{r}_{2}}{{(S{{O}_{4}})}_{3}} $
• D. Both (a) and (b)

Answer 1

Answer: (B)

$ {{K}_{2}}C{{r}_{2}}{{O}_{7}}+2{{H}_{2}}S{{O}_{4}}\xrightarrow{Cold} $ $ 2Cr{{O}_{3}}+2KHS{{O}_{4}}+{{H}_{2}}O $ $ Cr{{O}_{3}} $ also called chromic acid. It is highly acidic and oxidizing agent.