Question

monium phosphate ((NH4)3PO) is an important ingredient in many fertilizers. It can be made by reacting phosphoric acid (H3PO4) with ammonia (NH3) hat mass of ammonium phosphate is produced by the reaction of 5.55 g of phosphoric acid? sure your answer has the correct number of significant digits.

Answer 1

To determine the mass of ammonium phosphate \((NH_4)_3PO_4\) produced from the reaction of phosphoric acid (\(H_3PO_4\)) with ammonia \((NH_3)\), we can follow these steps:

1. **Write the balanced chemical equation:**

\[ H_3PO_4 + 3 NH_3 \rightarrow (NH_4)_3PO_4 \]

2. **Calculate the molar mass of phosphoric acid (H3PO4):**

- Hydrogen (H): \(1.01 \, \text{g/mol} \times 3 = 3.03 \, \text{g/mol}\)
- Phosphorus (P): \(30.97 \, \text{g/mol}\)
- Oxygen (O): \(16.00 \, \text{g/mol} \times 4 = 64.00 \, \text{g/mol}\)

\[ \text{Molar mass of } H_3PO_4 = 3.03 + 30.97 + 64.00 = 98.00 \, \text{g/mol} \]

3. **Convert mass of phosphoric acid to moles:**

\[ \text{Moles of } H_3PO_4 = \frac{\text{mass}}{\text{molar mass}} = \frac{5.55 \, \text{g}}{98.00 \, \text{g/mol}} \approx 0.0566 \, \text{mol} \]

4. **Determine moles of ammonium phosphate produced:**

According to the balanced equation, 1 mole of \(H_3PO_4\) produces 1 mole of \((NH_4)_3PO_4\).

\[ \text{Moles of } (NH_4)_3PO_4 = 0.0566 \, \text{mol} \]

5. **Calculate the molar mass of ammonium phosphate \((NH_4)_3PO_4\):**

- Nitrogen (N): \(14.01 \, \text{g/mol} \times 3 = 42.03 \, \text{g/mol}\)
- Hydrogen (H): \(1.01 \, \text{g/mol} \times 12 = 12.12 \, \text{g/mol}\)
- Phosphorus (P): \(30.97 \, \text{g/mol}\)
- Oxygen (O): \(16.00 \, \text{g/mol} \times 4 = 64.00 \, \text{g/mol}\)

\[ \text{Molar mass of } (NH_4)_3PO_4 = 42.03 + 12.12 + 30.97 + 64.00 = 149.12 \, \text{g/mol} \]

6. **Calculate the mass of ammonium phosphate produced:**

\[ \text{Mass of } (NH_4)_3PO_4 = \text{moles} \times \text{molar mass} = 0.0566 \, \text{mol} \times 149.12 \, \text{g/mol} \approx 8.44 \, \text{g} \]

**Final Answer:**

The mass of ammonium phosphate produced by the reaction of 5.55 g of phosphoric acid is \(8.44 \, \text{g}\). This answer is given with the correct number of significant digits, reflecting the three significant digits in the initial mass of phosphoric acid.