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:
Show Hint
- \( K_2Cr(OH)_6 \) and \( Cr_2O_3 \)
- \( K_2CrO_4 \) and \( Cr_2O_3 \)
- \( K_2CrO_4 \) and \( K_2Cr_2O_7 \)
- \( K_2CrO_4 \) and \( CrO \)
The Correct Option is C
Approach Solution - 1
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:
- 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.
- 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 \(K_2Cr_2O_7\)
Approach Solution -2
- 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] = \( K_2Cr_2O_7 \).
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