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To determine the mass of ammonium phosphate ((NH₄)₃PO₄) produced from 4.9 g of phosphoric acid (H₃PO₄), we first need to understand the balanced chemical equation for the reaction:
\[ \text{H₃PO₄} + 3 \text{NH₃} \rightarrow (NH₄)₃PO₄ \]
From the balanced equation, 1 mole of phosphoric acid (H₃PO₄) reacts with 3 moles of ammonia (NH₃) to produce 1 mole of ammonium phosphate ((NH₄)₃PO₄).
### Step-by-Step Calculation:
1. **Calculate the molar mass of phosphoric acid (H₃PO₄):**
- Hydrogen (H): 1.01 g/mol × 3 = 3.03 g/mol
- Phosphorus (P): 30.97 g/mol
- Oxygen (O): 16.00 g/mol × 4 = 64.00 g/mol
Total molar mass of H₃PO₄:
\[ 3.03 \, \text{g/mol} + 30.97 \, \text{g/mol} + 64.00 \, \text{g/mol} = 98.00 \, \text{g/mol} \]
2. **Convert the given mass of phosphoric acid (H₃PO₄) to moles:**
\[ \text{Moles of H₃PO₄} = \frac{\text{Mass}}{\text{Molar mass}} = \frac{4.9 \, \text{g}}{98.00 \, \text{g/mol}} = 0.05 \, \text{mol} \]
3. **Determine the moles of (NH₄)₃PO₄ produced:**
- According to the balanced equation, 1 mole of H₃PO₄ produces 1 mole of (NH₄)₃PO₄.
\[ \text{Moles of } (NH₄)₃PO₄ = 0.05 \, \text{mol} \]
4. **Calculate the mass of (NH₄)₃PO₄ produced:**
- Now, calculate the mass using the molar mass of (NH₄)₃PO₄:
- Ammonium (NH₄): 18.04 g/mol
- Phosphorus (P): 30.97 g/mol
- Oxygen (O): 16.00 g/mol × 4 = 64.00 g/mol
Total molar mass of (NH₄)₃PO₄:
\[ 18.04 \, \text{g/mol} + 30.97 \, \text{g/mol} + 64.00 \, \text{g/mol} = 113.01 \, \text{g/mol} \]
\[ \text{Mass of } (NH₄)₃PO₄ = \text{Moles} \times \text{Molar mass} \]
\[ \text{Mass of } (NH₄)₃PO₄ = 0.05 \, \text{mol} \times 113.01 \, \text{g/mol} = 5.65 \, \text{g} \]
### Conclusion:
Therefore, the mass of ammonium phosphate ((NH₄)₃PO₄) produced from 4.9 g of phosphoric acid (H₃PO₄) is **5.65 g**. This calculation considers the stoichiometry of the reaction and the molar masses of the substances involved.
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