Solvi Ammonium Phosphate ((NH4)3PO4) is an important ingredient in many fertilizers. It can be made by reacting phosphoric acid (H3PO4) with ammonia (NH3). What mass of ammonium phosphate is produced by the reaction of 4.6 g of phosphoric acid? Round your answer to 2 significant digits.
The Correct Answer and Explanation is:
To determine the mass of ammonium phosphate ((NH₄)₃PO₄) produced from 4.6 g of phosphoric acid (H₃PO₄), we follow these steps:
Step 1: Write the balanced chemical equation
H₃PO₄+3NH₃→(NH₄)3PO₄\text{H₃PO₄} + 3\text{NH₃} \rightarrow (\text{NH₄})₃\text{PO₄}
This reaction shows that 1 mole of H₃PO₄ reacts with 3 moles of NH₃ to form 1 mole of (NH₄)₃PO₄.
Step 2: Calculate the molar mass of H₃PO₄
- H: 1.01 × 3 = 3.03 g/mol
- P: 30.97 g/mol
- O: 16.00 × 4 = 64.00 g/mol
Total = 3.03 + 30.97 + 64.00 = 98.00 g/mol
Step 3: Convert 4.6 g H₃PO₄ to moles
4.6 g98.00 g/mol=0.04694 mol\frac{4.6 \text{ g}}{98.00 \text{ g/mol}} = 0.04694 \text{ mol}
Step 4: Use the mole ratio
From the balanced equation, 1 mol H₃PO₄ → 1 mol (NH₄)₃PO₄, so: 0.04694 mol H₃PO₄→0.04694 mol (NH₄)₃PO₄0.04694 \text{ mol H₃PO₄} → 0.04694 \text{ mol (NH₄)₃PO₄}
Step 5: Calculate molar mass of (NH₄)₃PO₄
- N: 14.01 × 3 = 42.03
- H: 1.01 × 12 = 12.12
- P: 30.97
- O: 16.00 × 4 = 64.00
Total = 42.03 + 12.12 + 30.97 + 64.00 = 149.12 g/mol
Step 6: Calculate mass of (NH₄)₃PO₄ produced
0.04694 mol×149.12 g/mol=6.999 g0.04694 \text{ mol} × 149.12 \text{ g/mol} = 6.999 \text{ g}
Final Answer:
7.0 g\boxed{7.0 \text{ g}}
Explanation
The production of ammonium phosphate ((NH₄)₃PO₄), a widely used fertilizer component, involves a reaction between phosphoric acid (H₃PO₄) and ammonia (NH₃). This is an example of an acid-base reaction, where phosphoric acid donates protons to form ammonium ions, which then combine with phosphate ions.
To determine how much ammonium phosphate is formed from 4.6 grams of H₃PO₄, we start with the balanced chemical equation. This equation shows a 1:1 molar ratio between phosphoric acid and ammonium phosphate, meaning each mole of H₃PO₄ yields one mole of (NH₄)₃PO₄, assuming excess ammonia is available.
Next, we calculate the molar mass of H₃PO₄, which is 98.00 g/mol. By dividing the given mass (4.6 g) by this molar mass, we find the number of moles of H₃PO₄: approximately 0.04694 mol.
Since the ratio is 1:1, the same number of moles of (NH₄)₃PO₄ will be produced. The next step is to calculate the molar mass of ammonium phosphate, which is 149.12 g/mol. Multiplying this by the number of moles gives us the final mass: around 6.999 g. Rounding this to two significant figures (based on the initial data), we obtain 7.0 g.
This calculation is important in both academic and industrial chemistry contexts, as precise stoichiometry is critical in producing the correct amount of product while minimizing waste and ensuring cost efficiency in large-scale fertilizer production.
