When 0.421 g of BaCl₂ · 2 H2O and 0.722g of Na₃PO₄ · 12 H₂O mix with water forming 500 mL of solution then 0.003 mol/L of Ba₃(PO4)₂ precipitate.
What is Stoichiometry ?
Stoichiometry is an important concept which helps us use the balanced chemical equation to measures the amount of reactants and products.
What is Molarity ?
Molarity of solution is the number of moles of solute per liter of solution. Molarity is also called Molar Concentration.
Molarity (M) = [tex]\frac{\text{number of moles}}{\text{volume of solution (in liters)}}[/tex]
How to find the number of moles ?
Number of moles = [tex]\frac{\text{Given mass}}{\text{Molar mass}}[/tex]
First we have to write the balanced chemical equation
2Na₃PO₄ · 12 H₂O + 3BaCl₂ · 2 H2O → Ba₃(PO4)₂ + 6 NaCl + 30 H₂O
Here 3 moles of BaCl₂ · 2 H2O produce 1 mole of Ba₃(PO4)₂
So,
Number of moles of BaCl₂ · 2 H2O = [tex]\frac{\text{Mass of}\ BaCl_2 . 2H_2O}{\text{Molar mass of}\ BaCl_2. 2H_2O}[/tex]
= [tex]\frac{0.421}{244.26}[/tex]
= 0.017
Number of moles of 2Na₃PO₄ · 12 H₂O = [tex]\frac{\text{Mass of}\ Na_3PO_4. 12H_2O}{\text{Molar mass of}\ Na_3PO_4 . 12H_2O}[/tex]
= [tex]\frac{0.722}{380.12}[/tex]
= 0.002
Therefore, Na₃PO₄ · 12 H₂O is the limiting reagent.
Now, put the value in above formula to find the molarity
Molarity (M) = [tex]\frac{\text{number of moles}}{\text{volume of solution (in liters)}}[/tex]
= [tex]\frac{0.0017}{0.5}[/tex]
= 0.003 mol/L
Thus, from above conclusion we can say that 0.003 mol/L of Ba₃(PO4)₂ precipitate.
Learn more about the Stoichiometry here: https://brainly.com/question/14935523
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