Dry hydrogen was passed over `1.58` g of red hot copper oxide till all of it completely reduced to 1.26 g of copper (Cu). If in this process 0.36 g of `H_2 O` is formed, what will be the equivalent weight of `Cu and O`?

Explanation

To solve the problem of finding the equivalent weight of copper (Cu) and oxygen (O) from the given data, we will follow these steps:

### Step 1: Write the Reaction
The reaction between copper oxide (CuO) and hydrogen (H₂) can be represented as:
\[ \text{CuO} + \text{H}_2 \rightarrow \text{Cu} + \text{H}_2\text{O} \]

### Step 2: Calculate the Moles of Water (H₂O) Formed
Given that 0.36 g of water is formed, we can calculate the number of moles of water using its molar mass (18 g/mol):
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