Looking for reliable answers? Westonci.ca is the ultimate Q&A platform where experts share their knowledge on various topics. Get detailed and precise answers to your questions from a dedicated community of experts on our Q&A platform. Get precise and detailed answers to your questions from a knowledgeable community of experts on our Q&A platform.
Sagot :
To determine the mass of ammonium phosphate produced from the reaction of 2.35 grams of phosphoric acid, we follow a series of steps involving stoichiometry and molar mass calculations. Here’s the step-by-step solution:
1. Identify the balanced chemical equation:
The balanced chemical equation for the reaction between phosphoric acid (H₃PO₄) and ammonia (NH₃) to form ammonium phosphate ((NH₄)₃PO₄) is:
[tex]\[ \text{H₃PO₄ + 3 NH₃ → (NH₄)₃PO₄} \][/tex]
2. Determine the molar masses:
- Molar mass of H₃PO₄ (phosphoric acid) = 98.00 g/mol
- Molar mass of (NH₄)₃PO₄ (ammonium phosphate) = 149.09 g/mol
3. Calculate the moles of phosphoric acid:
Given mass of phosphoric acid = 2.35 g
Moles of phosphoric acid can be found using the formula:
[tex]\[ \text{moles of H₃PO₄} = \frac{\text{mass of H₃PO₄}}{\text{molar mass of H₃PO₄}} \][/tex]
Substituting the known values:
[tex]\[ \text{moles of H₃PO₄} = \frac{2.35 \text{ g}}{98.00 \text{ g/mol}} \approx 0.02398 \text{ mol} \][/tex]
4. Use stoichiometry to find the moles of ammonium phosphate produced:
The balanced chemical equation shows that 1 mole of H₃PO₄ produces 1 mole of (NH₄)₃PO₄. Therefore, the moles of (NH₄)₃PO₄ produced are equal to the moles of H₃PO₄.
[tex]\[ \text{moles of (NH₄)₃PO₄} = 0.02398 \text{ mol} \][/tex]
5. Calculate the mass of ammonium phosphate produced:
Using the moles of ammonium phosphate and its molar mass:
[tex]\[ \text{mass of (NH₄)₃PO₄} = \text{moles of (NH₄)₃PO₄} \times \text{molar mass of (NH₄)₃PO₄} \][/tex]
Substituting the known values:
[tex]\[ \text{mass of (NH₄)₃PO₄} = 0.02398 \text{ mol} \times 149.09 \text{ g/mol} \approx 3.575 \text{ g} \][/tex]
Therefore, the mass of ammonium phosphate produced by the reaction of 2.35 grams of phosphoric acid, rounded to 3 significant digits, is:
[tex]\[ \boxed{3.575 \text{ g}} \][/tex]
1. Identify the balanced chemical equation:
The balanced chemical equation for the reaction between phosphoric acid (H₃PO₄) and ammonia (NH₃) to form ammonium phosphate ((NH₄)₃PO₄) is:
[tex]\[ \text{H₃PO₄ + 3 NH₃ → (NH₄)₃PO₄} \][/tex]
2. Determine the molar masses:
- Molar mass of H₃PO₄ (phosphoric acid) = 98.00 g/mol
- Molar mass of (NH₄)₃PO₄ (ammonium phosphate) = 149.09 g/mol
3. Calculate the moles of phosphoric acid:
Given mass of phosphoric acid = 2.35 g
Moles of phosphoric acid can be found using the formula:
[tex]\[ \text{moles of H₃PO₄} = \frac{\text{mass of H₃PO₄}}{\text{molar mass of H₃PO₄}} \][/tex]
Substituting the known values:
[tex]\[ \text{moles of H₃PO₄} = \frac{2.35 \text{ g}}{98.00 \text{ g/mol}} \approx 0.02398 \text{ mol} \][/tex]
4. Use stoichiometry to find the moles of ammonium phosphate produced:
The balanced chemical equation shows that 1 mole of H₃PO₄ produces 1 mole of (NH₄)₃PO₄. Therefore, the moles of (NH₄)₃PO₄ produced are equal to the moles of H₃PO₄.
[tex]\[ \text{moles of (NH₄)₃PO₄} = 0.02398 \text{ mol} \][/tex]
5. Calculate the mass of ammonium phosphate produced:
Using the moles of ammonium phosphate and its molar mass:
[tex]\[ \text{mass of (NH₄)₃PO₄} = \text{moles of (NH₄)₃PO₄} \times \text{molar mass of (NH₄)₃PO₄} \][/tex]
Substituting the known values:
[tex]\[ \text{mass of (NH₄)₃PO₄} = 0.02398 \text{ mol} \times 149.09 \text{ g/mol} \approx 3.575 \text{ g} \][/tex]
Therefore, the mass of ammonium phosphate produced by the reaction of 2.35 grams of phosphoric acid, rounded to 3 significant digits, is:
[tex]\[ \boxed{3.575 \text{ g}} \][/tex]
We hope you found this helpful. Feel free to come back anytime for more accurate answers and updated information. Your visit means a lot to us. Don't hesitate to return for more reliable answers to any questions you may have. Stay curious and keep coming back to Westonci.ca for answers to all your burning questions.