Respuesta :
Answer:
1.02 grams
Explanation:
The reaction between Na₃PO₄ and BaCl₂ is:
2Na₃PO₄(aq) + 3BaCl₂(aq) → Ba₃(PO₄)₂(s) + 6NaCl(aq)
We need to find the number of moles of Na₃PO₄ and BaCl₂:
[tex] n_{Na_{3}PO_{4}} = C*V = 0.200 M*0.050 L = 0.01 moles [/tex]
[tex] n_{BaCl_{2}} = C*V = 0.100 M*0.050 L = 0.005 moles [/tex]
Now, we need to find the limiting reactant:
[tex] n_{Na_{3}PO_{4}} = \frac{2}{3}*0.005 moles = 0.0033 moles [/tex]
We have that it is needed 0.0033 moles of Na₃PO₄ to react with BaCl₂ and we have 0.01 moles of Na₃PO₄, so the limiting reactant is BaCl₂.
The number of moles of Ba₃(PO₄)₂ is:
[tex] n_{Ba_{3}(PO_{4})_{2}} = \frac{0.005}{3} = 0.0017 moles [/tex]
Finally, the mass of Ba₃(PO₄)₂ is:
[tex] m = n_{Ba_{3}(PO_{4})_{2}}*M = 0.0017 moles*601.93 g/mol = 1.02 g [/tex]
Therefore, should be obtained 1.02 grams of the precipitate.
I hope it helps you!
The mass of the precipitate that should be obtained is 1.005 g
First, we will write the balanced chemical equation for the reaction.
The balanced chemical equation for the reaction is
2Na₃PO₄ + 3BaCl₂ → Ba₃(PO₄)₂ + 6NaCl
This means 2 moles of Na₃PO₄ react with 3 moles of BaCl₂ to produce 1 mole of Ba₃(PO₄)₂ and 6 moles of NaCl
Now, we will determine the number of moles of each reactant present
For Na₃PO₄
Volume = 50.0 mL = 0.050 L
Concentration = 0.200 M
From the formula
Number of moles = Concentration × Volume
∴ Number of moles of Na₃PO₄ present = 0.200 × 0.050
Number of moles of Na₃PO₄ present = 0.01 mole
For BaCl₂
Volume = 50.0 mL = 0.050 L
Concentration = 0.100 M
∴ Number of moles of BaCl₂ present = 0.100 × 0.050
Number of moles of BaCl₂ present = 0.005 mole
Since
2 moles of Na₃PO₄ react with 3 moles of BaCl₂ to produce 1 mole of Ba₃(PO₄)₂
Then,
[tex]\frac{0.005\times 2}{3}[/tex] mole of Na₃PO₄ react with 0.005 mole of BaCl₂ to produce [tex]\frac{0.005}{3}[/tex] mole of Ba₃(PO₄)₂
That is,
0.0033 mole of Na₃PO₄ react with 0.005 mole of BaCl₂ to produce 0.00167 mole of Ba₃(PO₄)₂
∴ Number of moles of the precipitate [Ba₃(PO₄)₂] formed is 0.00167 mole
Now, for the mass of the precipitate that should be formed
Using the formula
Mass = Number of moles × Molar mass
Molar mass of Ba₃(PO₄)₂ = 601.93 g/mol
∴ Mass of the precipitate that should be produced = 0.00167 × 601.93
Mass of the precipitate that should be produced = 1.005 g
Hence, the mass of the precipitate that should be obtained is 1.005 g
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