The Ba3(PO4)2 (molar mass = 602.0 g/mol) precipitate that formed from a salt mixture has a mass of 0.767 g. If the Na3PO4·12H2O (molar mass = 380.2 g/mol) was the limiting reactant and BaCl2·2H2O was the excess reactant in the mixture, what is the mass of Na3FPO4·12H2O in the mixture? Select one: a) 0.812 g b) 0.690 g c) 0.716 g d) 0.968 g e) 0.933 g
The Correct Answer and Explanation is:
To find the mass of Na₃PO₄·12H₂O (sodium phosphate dodecahydrate) in the mixture, we use stoichiometry based on the information provided:
Given:
- Mass of Ba₃(PO₄)₂ precipitate = 0.767 g
- Molar mass of Ba₃(PO₄)₂ = 602.0 g/mol
- Molar mass of Na₃PO₄·12H₂O = 380.2 g/mol
- Na₃PO₄·12H₂O is the limiting reactant
- BaCl₂·2H₂O is in excess
Step 1: Moles of Ba₃(PO₄)₂ formed
Moles of Ba₃(PO₄)₂=0.767 g602.0 g/mol=0.001275 mol\text{Moles of Ba₃(PO₄)₂} = \frac{0.767\ \text{g}}{602.0\ \text{g/mol}} = 0.001275\ \text{mol}Moles of Ba₃(PO₄)₂=602.0 g/mol0.767 g=0.001275 mol
Step 2: Mole Ratio Between Ba₃(PO₄)₂ and Na₃PO₄
From the balanced chemical equation:3BaCl2+2Na3PO4→Ba3(PO4)2↓+6NaCl3BaCl₂ + 2Na₃PO₄ → Ba₃(PO₄)₂↓ + 6NaCl3BaCl2+2Na3PO4→Ba3(PO4)2↓+6NaCl
So:
- 2 moles of Na₃PO₄ produce 1 mole of Ba₃(PO₄)₂
Moles of Na₃PO₄ used=2×0.001275=0.002550 mol\text{Moles of Na₃PO₄ used} = 2 × 0.001275 = 0.002550\ \text{mol}Moles of Na₃PO₄ used=2×0.001275=0.002550 mol
Step 3: Mass of Na₃PO₄·12H₂O in the mixture
Mass=0.002550 mol×380.2 g/mol=0.9695 g\text{Mass} = 0.002550\ \text{mol} × 380.2\ \text{g/mol} = 0.9695\ \text{g}Mass=0.002550 mol×380.2 g/mol=0.9695 g
Rounded to three significant figures:0.968 g\boxed{0.968\ \text{g}}0.968 g
Correct Answer: d) 0.968 g
Explanation
This problem involves using stoichiometry to determine the mass of a reactant based on the amount of precipitate formed. We are told that barium phosphate, Ba₃(PO₄)₂, precipitates from a reaction between barium chloride and sodium phosphate. We also know the mass of the precipitate formed and that sodium phosphate is the limiting reactant, meaning it determines the amount of product made.
To begin, we convert the given mass of barium phosphate to moles using its molar mass. This gives us the amount of product in moles. According to the balanced equation, two moles of sodium phosphate react to form one mole of barium phosphate. So, we multiply the moles of Ba₃(PO₄)₂ by 2 to find the moles of sodium phosphate that were used in the reaction.
After calculating the moles of sodium phosphate, we convert it back to mass. Since the sodium phosphate is hydrated (with 12 water molecules), we use the molar mass of Na₃PO₄·12H₂O in our calculation. Multiplying the moles by this molar mass gives the total mass of sodium phosphate that reacted, which is also the amount that was present in the mixture because it was the limiting reactant.
This process demonstrates how stoichiometry helps relate the mass of reactants and products using mole ratios and molar masses. The final answer, based on careful calculations and proper rounding, is 0.968 grams of sodium phosphate dodecahydrate, corresponding to option (d).
