Question

Consider the following reactions: \[ K_2Cr_2O_7 + 2 KOH \xrightarrow{ \text{heat} } [A] \] \[ [A] + H_2SO_4 \xrightarrow{} [B] + K_2SO_4 \] The products [A] and [B], respectively are:

• A. \( K_2Cr(OH)_6 \) and \( Cr_2O_3 \)
• B. \( K_2CrO_4 \) and \( Cr_2O_3 \)
• C. \( K_2CrO_4 \) and \( K_2Cr_2O_7 \)
• D. \( K_2CrO_4 \) and \( CrO \)

Hint:

When potassium chromate reacts with sulfuric acid, it produces chromium trioxide. The reaction typically forms an orange solution due to the presence of \( K_2Cr_2O_7 \).

Answer 1

The Correct Option is C

To solve this question, we need to analyze the given chemical reactions step-by-step and determine the products [A] and [B]. Let's go through the reactions:

1. The first reaction is:

\(K_2Cr_2O_7 + 2 KOH \xrightarrow{ \text{heat} } [A]\)

In this reaction, potassium dichromate \(K_2Cr_2O_7\) reacts with potassium hydroxide \(KOH\) under heat. The product formed, [A], is potassium chromate \(K_2CrO_4\). This involves a change from the dichromate anion to the chromate anion.

1. The second reaction is:

\([A] + H_2SO_4 \xrightarrow{} [B] + K_2SO_4\)

In this reaction, [A] (which we identified as \(K_2CrO_4\)) reacts with sulfuric acid \(H_2SO_4\). This reaction converts the chromate ion back to the dichromate ion. The product [B] is potassium dichromate, \(K_2Cr_2O_7\), and potassium sulfate \(K_2SO_4\) is also formed.

Therefore, the correct products for the reactions are:

• [A]: \(K_2CrO_4\)
• [B]: \(K_2Cr_2O_7\)

Given these deductions, the correctly matched products according to the options provided are:

\(K_2CrO_4\) and \(K_2Cr_2O_7\)

Answer 2

The first reaction shows the reaction of \( K_2Cr_2O_7 \) (potassium dichromate) with KOH (potassium hydroxide) under heat. The product formed is potassium chromate \( K_2CrO_4 \), which is the compound [A].
- In the second reaction, \( K_2CrO_4 \) reacts with sulfuric acid (H2SO4), resulting in chromium trioxide \( Cr_2O_3 \) as the product [B]. Thus, the correct products are [A] = \( K_2CrO_4 \) and [B] = \( K_2Cr_2O_7 \).