Which is cheaper
40% HCl at the rate of 6 per kg or 80% `H_(2)SO_(4)` at the rate of 3.5 per kg required to neutralise 7 kg of KOH.

Explanation

To determine which is cheaper between 40% HCl at the rate of 6 per kg and 80% H₂SO₄ at the rate of 3.5 per kg required to neutralize 7 kg of KOH, we will follow these steps:

### Step 1: Write the neutralization reactions
1. **HCl with KOH**:
\[ \text{HCl} + \text{KOH} \rightarrow \text{KCl} + \text{H}_2\text{O} \]
- Stoichiometry: 1 mole of HCl reacts with 1 mole of KOH.
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