Final answer:
To calculate the quantity in moles of precipitate formed in the reaction between NaI and Pb(NO₃)₂, we must determine the limiting reactant, which in this case is Pb(NO₃)₂. Based on the stoichiometry of the reaction, 0.090 moles of Pb(NO₃)₂ will produce an equal amount of PbI₂ precipitate: 0.090 moles.
Step-by-step explanation:
The quantity in moles of precipitate that will be formed when 68.0 mL of 2.00 M NaI is reacted with 100.0 mL of 0.900 M Pb(NO₃)₂ can be determined by a stoichiometry calculation based on the balanced chemical reaction:
2 NaI (aq) + Pb(NO₃)₂ (aq) → PbI₂ (s) + 2 NaNO₃ (aq)
First, calculate the moles of NaI and Pb(NO₃)₂:
Moles of NaI = 0.068 L × 2.00 M = 0.136 moles
Moles of Pb(NO₃)₂ = 0.100 L × 0.900 M = 0.090 moles
According to the stoichiometry of the reaction, 1 mole of Pb(NO₃)₂ reacts with 2 moles of NaI to produce 1 mole of PbI₂. Thus, Pb(NO₃)₂ is the limiting reagent.
Since 1 mole of Pb(NO₃)₂ produces 1 mole of PbI₂, the moles of PbI₂ precipitate will be equal to the moles of Pb(NO₃)₂:
Moles of PbI₂ precipitate = Moles of Pb(NO₃)₂ = 0.090 moles