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]
