Respuesta :
Answer:
0.10M of Ba²⁺ is the concentration of the metal in excess
Explanation:
Based on the chemical reaction:
K₂CO₃(aq) + Ba(NO₃)₂(aq) → BaCO₃(s) + 2KNO₃(aq)
1 mole of potassium carbonate reacts per mole of barium nitrate
To solve this question we need to find the moles of each salt to find then the moles of Barium in excess:
Moles K₂CO₃:
0.025L * (0.25mol / L) = 0.00625moles K₂CO₃ = moles CO₃²⁻
Moles Ba(NO₃)₂:
0.030L * (0.40mol/L) = 0.012 moles of Ba(NO₃)₂ = 0.012 moles of Ba²⁺
That means moles of Ba²⁺ that don't react are:
0.012 mol - 0.00625mol = 0.00575 moles Ba²⁺
In 25 + 30mL = 55mL:
0.00575 moles Ba²⁺ / 0.055L =
0.10M of Ba²⁺ is the concentration of the metal in excess
The concentration of the excess metal ion after the precipitation reaction is complete is 0.1045 M
From the question,
We are to determine the concentration of the excess metal ion after the precipitation reaction is complete
First, we will write the balanced chemical equation for the reaction
The balanced chemical equation for the reaction is
K₂CO₃ + Ba(NO₃)₂ → BaCO₃ + 2KNO₃
This means
1 mole of K₂CO₃ is required to react completely with 1 mole of Ba(NO₃)₂
Now, we will determine the number of moles of each reactant present.
- For K₂CO₃
Volume = 25.0 mL = 0.025 L
Concentration = 0.25 M
Using the formula
Number of moles = Concentration × Volume
∴ Number of moles of K₂CO₃ present = 0.25 × 0.025
Number of moles of K₂CO₃ present = 0.00625 mole
- For Ba(NO₃)₂
Volume = 30.0 mL = 0.03 L
Concentration = 0.40 M
∴ Number of moles of Ba(NO₃)₂ present = 0.40 × 0.03
Number of moles of Ba(NO₃)₂ present = 0.012 mole
Since
1 mole of K₂CO₃ is required to react completely with 1 mole of Ba(NO₃)₂
Then,
The 0.00625 mole of K₂CO₃ will react with 0.00625 mole of Ba(NO₃)₂
∴ Only 0.00625 mole of the Ba(NO₃)₂ will react
This means Ba(NO₃)₂ is the excess reactant.
Now, we will determine the number of moles of Ba(NO₃)₂ remaining
Number of moles of Ba(NO₃)₂ remaining = 0.012 mole - 0.00625 mole
Number of moles of Ba(NO₃)₂ remaining = 0.00575 mole
Now, for the concentration
Using the formula
[tex]Concentration = \frac{Number\ of\ moles}{Volume}[/tex]
Volume = Total volume of the solution = 0.025L + 0.03L
Volume = 0.055 L
∴ Concentration of the excess metal ion = [tex]\frac{0.00575}{0.055}[/tex]
Concentration of the excess metal ion = 0.1045 M
Hence, the concentration of the excess metal ion after the precipitation reaction is complete is 0.1045 M
Learn more here: https://brainly.com/question/1825674
