According to the following reaction, what mass of PbCl2 can form from 235 mL of 0.110 M KCl solution? Assume that there is excess Pb(NO3)2. 2 KCl(aq) + Pb(NO3)2(aq) → PbCl2(s) + 2 KNO3(aq)

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:

Molarity is expressed in units

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?

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?

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?

mass=3.59 g

3.59 g of PbCl₂ can form from 235 mL of 0.110 M KCl solution.

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explanation: