Final answer:
When 291.0 mL of 0.350 M NaCl is reacted with excess Pb(NO₃)₂, 0.050925 moles of PbCl₂ precipitate will be formed, based on the stoichiometry of the reaction.
Step-by-step explanation:
To calculate the quantity in moles of precipitate formed when 291.0 mL of 0.350 M NaCl is reacted with excess Pb(NO₃)₂, we use the stoichiometry of the balanced chemical reaction:
Pb(NO₃)₂(aq) + 2 NaCl(aq) → PbCl₂(s) + 2 NaNO₃(aq)
According to this reaction, one mole of Pb(NO₃)₂ reacts with two moles of NaCl to form one mole of PbCl₂ precipitate. First, we need to convert the volume of NaCl solution to moles:
(0.291 L) × (0.350 mol/L) = 0.10185 mol NaCl
Using the reaction stoichiometry, we see that 1 mole of Pb(NO₃)₂ is needed for every 2 moles of NaCl. Thus, the moles of PbCl₂ precipitate formed from 0.10185 mol of NaCl are:
(0.10185 mol NaCl) × (1 mol PbCl₂ / 2 mol NaCl) = 0.050925 mol PbCl₂
Therefore, 0.050925 moles of PbCl₂ precipitate will be formed from the reaction of 291 ml of 0.350 M NaCl solution with excess Pb(NO₃)₂.