To determine how much precipitate is produced, we need to use stoichiometry and the concept of limiting reactants.
The balanced chemical equation for the reaction between lead(II) nitrate (Pb(NO3)2) and potassium chloride (KCl) is:
Pb(NO3)2 + 2KCl -> PbCl2 + 2KNO3
From the equation, we can see that 1 mole of lead(II) nitrate reacts with 2 moles of potassium chloride to produce 1 mole of lead(II) chloride and 2 moles of potassium nitrate.
First, we need to determine the number of moles of lead(II) nitrate:
Molarity = moles/volume
0.150 M = moles/35.0 L
moles of Pb(NO3)2 = 0.150 M x 35.0 L = 5.25 moles
According to the stoichiometry of the balanced equation, this means that we need 5.25 moles of potassium chloride. However, the problem states that we have an excess amount of potassium chloride, so the lead(II) nitrate is the limiting reactant.
Since 1 mole of lead(II) nitrate produces 1 mole of lead(II) chloride, we expect 5.25 moles of lead(II) chloride to be produced. However, we still need to convert moles to grams of lead(II) chloride:
Molar mass of PbCl2 = 207.2 g/mol
Mass of PbCl2 = 5.25 moles x 207.2 g/mol = 1086.6 g
Therefore, approximately 1086.6 grams of precipitate (lead(II) chloride) will be produced.
35.0 of 0.150 M lead 2 nitrate and excess amount of potassium chloride are mixed. How much precipitate is produced?
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