When 0.629 grams of Na3PO4.12H2O and 0.527 g BaCl2.2H2O are mixed with water , how many grams of Ba3(PO4)2 are produced?
The correct answer and explanation is :
When 0.629 grams of sodium phosphate dodecahydrate (Na₃PO₄·12H₂O) and 0.527 grams of barium chloride dihydrate (BaCl₂·2H₂O) are mixed in water, barium phosphate (Ba₃(PO₄)₂) precipitates out. To determine the mass of Ba₃(PO₄)₂ produced, follow these steps:
1. Write the balanced chemical equation:
[ 3 \, \text{Ba}^{2+} + 2 \, \text{PO}_4^{3-} \rightarrow \text{Ba}_3(\text{PO}_4)_2 ]
This equation shows that three barium ions react with two phosphate ions to form one unit of barium phosphate.
2. Calculate the molar masses of the reactants:
- Na₃PO₄·12H₂O: Sodium phosphate dodecahydrate has a molar mass of approximately 380.12 g/mol.
- BaCl₂·2H₂O: Barium chloride dihydrate has a molar mass of approximately 244.26 g/mol.
3. Determine the moles of each reactant:
- Na₃PO₄·12H₂O:
[ \frac{0.629 \, \text{g}}{380.12 \, \text{g/mol}} \approx 0.001658 \, \text{mol} ]
- BaCl₂·2H₂O:
[ \frac{0.527 \, \text{g}}{244.26 \, \text{g/mol}} \approx 0.002160 \, \text{mol} ]
4. Identify the limiting reactant:
From the balanced equation, the molar ratio of Ba²⁺ to PO₄³⁻ is 3:2. Therefore, for every 3 moles of Ba²⁺, 2 moles of PO₄³⁻ are required. Using the moles calculated:
- Ba²⁺: 0.002160 mol
- PO₄³⁻: 0.001658 mol × (2/3) ≈ 0.001105 mol
Since 0.001105 mol of PO₄³⁻ is less than 0.001658 mol of Ba²⁺, phosphate ions are the limiting reactant.
5. Calculate the moles of Ba₃(PO₄)₂ produced:
From the stoichiometry of the reaction, 2 moles of PO₄³⁻ produce 1 mole of Ba₃(PO₄)₂. Therefore:
[ 0.001105 \, \text{mol} \, \text{PO}_4^{3-} \times \frac{1 \, \text{mol} \, \text{Ba}_3(\text{PO}_4)_2}{2 \, \text{mol} \, \text{PO}_4^{3-}} \approx 0.000553 \, \text{mol} \, \text{Ba}_3(\text{PO}_4)_2 ]
6. Calculate the mass of Ba₃(PO₄)₂ produced:
The molar mass of Ba₃(PO₄)₂ is approximately 601.93 g/mol. Thus:
[ 0.000553 \, \text{mol} \times 601.93 \, \text{g/mol} \approx 0.333 \, \text{g} ]
Conclusion:
Approximately 0.333 grams of Ba₃(PO₄)₂ are produced when 0.629 grams of Na₃PO₄·12H₂O and 0.527 grams of BaCl₂·2H₂O are mixed with water.
This calculation is based on the stoichiometric relationships between the reactants and products, considering the limiting reactant and the molar masses involved.