(a). What is the equivalent weight of (i) `Fe(HC_2O_4)_2`

Explanation

To find the equivalent weight of \( \text{Fe(HC}_2\text{O}_4)_2 \), we will follow these steps:

### Step 1: Determine the molecular weight of \( \text{Fe(HC}_2\text{O}_4)_2 \)

1. **Identify the components of the compound**:
- Iron (Fe): Atomic weight = 55.85 g/mol
- Oxalic acid ion (HC₂O₄): Molecular weight = 1 (H) + 2(12.01) (C) + 4(16.00) (O) = 1 + 24.02 + 64.00 = 89.02 g/mol
- Since there are 2 HC₂O₄ ions, the total weight contributed by oxalic acid = 2 × 89.02 = 178.04 g/mol
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