A precipitation reaction occurs when 50. 0 mL of 0. 50 M BaCl₂ is mixed with 75. 0 mL of 0. 75 M Na₂CO₃ (aq). The only precipitate is the BaCO₃ (s) formed. (a) The balanced chemical equation is BaCl₂ + Na₂CO₃ → BaCO₃ + 2NaCl. (b) The limiting reagent is Na₂CO₃.
The balanced chemical equation is the equation in which the number of atoms on the reactant side is equal to the number of atoms on the product side in an equation.
Now we have to write the balanced equation
BaCl₂ + Na₂CO₃ → BaCO₃ + 2NaCl
To calculate number of moles when molarity and volume are given as:
Number of moles = Molarity × Volume
Now,
Number of moles of BaCl₂ = Molarity × Volume
= 0.50 M × 0.05 L [1 ml = 0.001 L]
= 0.025 moles
Number of moles of Na₂CO₃ = Molarity × Volume
= 0.75 M × 0.075 L [1 ml = 0.001 L]
= 0.05625 moles
The limiting reagent is Na₂CO₃ here.
Thus from the above conclusion we can say that A precipitation reaction occurs when 50. 0 mL of 0. 50 M BaCl₂ is mixed with 75. 0 mL of 0. 75 M Na₂CO₃ (aq). The only precipitate is the BaCO₃ (s) formed. (a) The balanced chemical equation is BaCl₂ + Na₂CO₃ → BaCO₃ + 2NaCl. (b) The limiting reagent is Na₂CO₃.
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Question: A precipitation reaction occurs when 50. 0 mL of 0. 50 M BaCl₂ is mixed with 75. 0 mL of 0. 75 M Na₂CO₃ (aq). The only precipitate is the BaCO₃ (s) formed.
a) Write the balanced equation that describes this reaction.
b) Which chemical is the limiting reactant
