Respuesta :
Answer:
(B) 0.160 M
Explanation:
Considering:
[tex]Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}[/tex]
Or,
[tex]Moles =Molarity \times {Volume\ of\ the\ solution}[/tex]
Given :
For [tex]K_2CO_3[/tex] :
Molarity = 0.200 M
Volume = 20.0 mL
The conversion of mL to L is shown below:
1 mL = 10⁻³ L
Thus, volume = 20.0×10⁻³ L
Thus, moles of [tex]K_2CO_3[/tex] :
[tex]Moles=0.200 \times {20.0\times 10^{-3}}\ moles[/tex]
Moles of [tex]K_2CO_3[/tex] = 0.004 moles
For [tex]Ba(NO_3)_2[/tex] :
Molarity = 0.400 M
Volume = 30.0 mL
The conversion of mL to L is shown below:
1 mL = 10⁻³ L
Thus, volume =30.0×10⁻³ L
Thus, moles of [tex]Ba(NO_3)_2[/tex] :
[tex]Moles=0.400\times {30.0\times 10^{-3}}\ moles[/tex]
Moles of [tex]Ba(NO_3)_2[/tex] = 0.012 moles
According to the given reaction:
[tex]K_2CO_3_{(aq)}+Ba(NO_3)_2_{(aq)}\rightarrow BaCO_3_{(s)}+2KNO_3_{(aq)}[/tex]
1 mole of potassium carbonate react with 1 mole of barium nitrate
0.004 moles potassium carbonate react with 0.004 mole of barium nitrate
Moles of barium nitrate = 0.004 moles
Available moles of barium nitrate = 0.012 moles
Limiting reagent is the one which is present in small amount. Thus, potassium carbonate is limiting reagent.
The formation of the product is governed by the limiting reagent. So,
1 mole of potassium carbonate gives 1 mole of barium carbonate
Also,
0.004 mole of potassium carbonate gives 0.004 mole of barium carbonate
Mole of barium carbonate = 0.004 moles
Also, consumed barium nitrate = 0.004 moles (barium ions precipitate with carbonate ions)
Left over moles = 0.012 - 0.004 moles = 0.008 moles
Total volume = 20.0 + 30.0 mL = 50.0 mL = 0.05 L
So, Concentration = 0.008/0.05 M = 0.160 M
(B) is correct.
The molarity of the barium ion in the solution has been 0.16 M. Thus, option B is correct.
The balanced reaction can be given as:
[tex]\rm K_2CO_3\;+\;Ba(NO_3)_2\;\rightarrow\;Ba_2CO_3\;+\;2\;KNO_3[/tex]
The molarity can be defined as moles of solute present in a liter of solution.
Molarity = [tex]\rm \dfrac{moles}{Volume\;(L)}[/tex]
Moles = Molarity × Volume (L)
- The moles of potassium carbonate in 20ml 0.2M solution has been:
Moles of Potassium carbonate = 0.2mol/L [tex]\rm \times[/tex] 0.02 L
Moles of Potassium carbonate = 0.004 g/mol
- The moles of Barium nitrate in 30ml 0.4M solution has been:
Moles of Barium nitrate = 0.4mol/L [tex]\rm \times[/tex] 0.03 L
Moles of Barium nitrate = 0.012 g/mol
From the balanced equation, 1 moles potassium carbonate reacts with 1 moles of barium nitrate.
1 mole Potassium carbonate = 1 mole Barium nitrate
0.004 Potassium nitrate = 0.004 mol Barium nitrate.
Since Barium nitrate has been present in excess than required, Potassium carbonate has been the limiting reactant.
After the reaction, the Barium carbonate left in the solution will provide barium ions.
The barium ions left in the solution will be:
= Moles of Barium Carbonate = Consumed barium carbonate
= 0.012 mol - 0.004 mol
= 0.008 mol.
The moles of Barium ions left in the solution have been 0.008 mol. The total volume of the solution after the reaction has been:
Total volume = Volume of Potassium carbonate + Volume of Barium nitrate
Total volume = 30 ml + 20 ml
Total volume = 50 ml.
The concentration of the Barium ions in the solution can be expressed in terms of molarity.
The molarity of the Barium ions has been:
Molarity = [tex]\rm \dfrac{moles}{Volume\;(L)}[/tex]
Molarity of Barium ions = [tex]\rm \dfrac{0.008\;mol}{0.05\;L}[/tex]
Molarity of Barium ions = 0.16 M.
The molarity of the barium ion in the solution has been 0.16 M. Thus, option B is correct.
For more information about the concentration of ions, refer to the link:
https://brainly.com/question/3045247
