Answer:
3.59 g of PbCl₂ can form from 235 mL of 0.110 M KCl solution.
Explanation:
2 KCl(aq) + Pb(NO₃)₂(aq) → PbCl₂(s) + 2 KNO₃(aq)
The Molarity (M) or Molar Concentration is the number of moles of solute (substance that is in minor proportion in a solution) that are dissolved in a certain volume.
The Molarity of a solution is determined by the following formula:
[tex]Molarity M=\frac{number of moles of solute}{Volume}[/tex]
Molarity is expressed in units [tex]\frac{moles}{liter}[/tex]
Then, knowing that 235 mL is 0.235 L (1 L=1000 mL or 1 mL=0.001 L) and taking into account the molarity of KCl, you can apply a rule of three as follows: if in 1 L of KCl solution there are 0.110 moles of the substance, in 0.235 L of solution how many moles of KCl are there?
[tex]moles of KCl=\frac{0.235 L*0.110moles}{1L}[/tex]
moles of KCl=0.02585
Another rule of three is applied as follows: by stoichiometry of the reaction you can see that 2 moles of KCl form 1 mole of PbCl₂. So if 0.02585 moles of KCl react, how many moles of PbCl₂ are formed?
[tex]molesofPbCl_{2} =\frac{0.02585molesofKCl*1molofPbCl_{2} }{2molesofKCl}[/tex]
moles of PbCl₂=0.012925
In the periodic table of elements you can see the following atomic masses:
Then the molar mass of PbCl₂ is:
PbCl₂= 207 g/mol + 2*35.45 g/mol= 277.9 g/mol
Then a last rule of three applies: if in 1 mole of PbCl₂ there are 277.9 g of the compound, in 0.012925 moles of PbCl₂ how much mass is there?
[tex]mass=\frac{0.012925moles*277.9g}{1 mole}[/tex]
mass=3.59 g
3.59 g of PbCl₂ can form from 235 mL of 0.110 M KCl solution.
The mass of PbCl₂ obtained from the reaction between 235 mL of 0.110 M KCl solution and excess Pb(NO₃)₂ is 3.59 g
We'll begin by calculating the number of mole of KCl in the solution.
Volume of KCl solution = 235 mL = 235 / 1000 = 0.235 L
Molarity of KCl = 0.110 M
Mole = Molarity × Volume
Mole of KCl = 0.110 × 0.235
Next, we shall determine the number of mole of PbCl₂ produced from the reaction.
2KCl(aq) + Pb(NO₃)₂(aq) → PbCl₂(s) + 2KNO₃(aq)
From the balanced equation above,
2 moles of KCl reacted to produce 1 mole of PbCl₂.
Therefore,
0.02585 mole of KCl will react to produce = [tex]\frac{0.02585}{2} \\\\[/tex] = 0.012925 mole of PbCl₂.
Finally, we shall determine the mass of 0.0129 mole of PbCl₂. This can be obtained as follow:
Molar mass of PbCl₂ = 207 + (35.5×2) = 278 g/mol
Mole of PbCl₂ = 0.012925 mole
Mass = mole × molar mass
Mass of PbCl₂ = 0.012925 × 278
Therefore, the mass of PbCl₂ obtained from the reaction is 3.59 g
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