`50 g` of `CaCO_(3)` is allowed to react with `70 g` of `H_(3)PO_(4^(.)` Calculate :
`(i)` amount of `Ca_(3)(PO_(4))_(2)` formed
`(ii)` amount of unreacted reagent

Explanation

To solve the given problem, we will follow these steps:

### Step 1: Write the balanced chemical equation
The reaction between calcium carbonate (CaCO₃) and phosphoric acid (H₃PO₄) can be represented as follows:

\[ 3 \text{CaCO}_3 + 2 \text{H}_3\text{PO}_4 \rightarrow \text{Ca}_3(\text{PO}_4)_2 + 3 \text{H}_2\text{O} + 3 \text{CO}_2 \]

### Step 2: Calculate the number of moles of reactants
1. **For CaCO₃:**
- Molar mass of CaCO₃ = 40 (Ca) + 12 (C) + 3 × 16 (O) = 100 g/mol
- Moles of CaCO₃ = \(\frac{50 \text{ g}}{100 \text{ g/mol}} = 0.5 \text{ moles}\)

2. **For H₃PO₄:**
- Molar mass of H₃PO₄ = 3 (H) + 31 (P) + 4 × 16 (O) = 98 g/mol
- Moles of H₃PO₄ = \(\frac{70 \text{ g}}{98 \text{ g/mol}} \approx 0.714 \text{ moles}\)

### Step 3: Determine the limiting reactant
From the balanced equation, we see that:
- 3 moles of CaCO₃ react with 2 moles of H₃PO₄.

To find the required moles of H₃PO₄ for 0.5 moles of CaCO₃:
- Required moles of H₃PO₄ = \(\frac{2}{3} \times 0.5 = \frac{1}{3} \approx 0.333 \text{ moles}\)

Since we have 0.714 moles of H₃PO₄ available, and only 0.333 moles are required, CaCO₃ is the limiting reactant.

### Step 4: Calculate the amount of Ca₃(PO₄)₂ formed
From the balanced equation:
- 3 moles of CaCO₃ produce 1 mole of Ca₃(PO₄)₂.

Thus, 0.5 moles of CaCO₃ will produce:
- Moles of Ca₃(PO₄)₂ = \(\frac{0.5}{3} = \frac{1}{6} \approx 0.167 \text{ moles}\)

### Step 5: Calculate the mass of Ca₃(PO₄)₂ formed
- Molar mass of Ca₃(PO₄)₂ = 3 × 40 (Ca) + 2 × (31 (P) + 4 × 16 (O)) = 310 g/mol
- Mass of Ca₃(PO₄)₂ = \(0.167 \text{ moles} \times 310 \text{ g/mol} \approx 51.67 \text{ g}\)

### Step 6: Calculate the amount of unreacted H₃PO₄
From the balanced equation:
- 2 moles of H₃PO₄ are required for 3 moles of CaCO₃.
- For 0.5 moles of CaCO₃, required moles of H₃PO₄ = \(\frac{2}{3} \times 0.5 = \frac{1}{3} \approx 0.333 \text{ moles}\)

Unreacted moles of H₃PO₄:
- Unreacted moles = Initial moles - Required moles = \(0.714 - 0.333 = 0.381 \text{ moles}\)

### Step 7: Calculate the mass of unreacted H₃PO₄
- Mass of unreacted H₃PO₄ = \(0.381 \text{ moles} \times 98 \text{ g/mol} \approx 37.33 \text{ g}\)

### Final Answers:
(i) Amount of Ca₃(PO₄)₂ formed = **51.67 g**
(ii) Amount of unreacted H₃PO₄ = **37.33 g**