Question

Malachite decomposed to give \(A + CO_2 + H_2O\) and compound \(A\) on reduction with carbon gives \(CO + B\). Here, \(A\) and \(B\) are

• A. CuO, Cu
• B. Cu\(_2\)O, CuO
• C. Cu\(_2\)O, Cu
• D. CuCO\(_3\), Cu

Hint:

Malachite on heating gives oxide, which on reduction gives metal.

Answer 1

The Correct Option is C

To solve the problem of identifying the decomposition and reduction products of malachite, we need to understand the chemical reactions involved.

1. Malachite is a basic copper carbonate mineral with the formula \(CuCO_3 \cdot Cu(OH)_2\). When it decomposes upon heating, it breaks down into copper(I) oxide \((Cu_2O)\), carbon dioxide, and water.
2. The decomposition reaction can be represented as follows: \(2[CuCO_3 \cdot Cu(OH)_2] \rightarrow 2Cu_2O + 2CO_2 + 2H_2O\).
3. According to the given question, after decomposition, the compound \(A\) is the residue from decomposition, which is copper(I) oxide \((Cu_2O)\).
4. Compound \(A\) (\(Cu_2O\)) is then reduced using carbon, according to the following reaction: \(Cu_2O + C \rightarrow 2Cu + CO\). This reaction produces carbon monoxide \((CO)\) and copper metal \((Cu)\).
5. Therefore, compound \(B\), formed from the reduction of compound \(A\) with carbon, is copper metal \((Cu)\).

Thus, the correct answer is:

Cu₂O, Cu
 

This matches with the given options and confirms that compound \(A\) is \(Cu_2O\) and compound \(B\) is \(Cu\).