Final answer:
To determine the result of mixing Na₃PO₄ and Pb(NO₃)₂ solutions, calculate the moles of each reactant and identify the limiting reactant based on the stoichiometric ratios. We find Pb(NO₃)₂ to be the limiting reactant, which dictates the quantity of Pb₃(PO₄)₂ formed in the reaction.
Step-by-step explanation:
To determine the outcome of the reaction between a solution of Na₃PO₄ and a solution of Pb(NO₃)₂, we need to consider the stoichiometry of the balanced chemical equation provided: 2Na₃PO₄ (aq) + 3Pb(NO₃)₂ (aq) → Pb₃(PO₄)₂ (s) + 6NaNO₃ (aq).
Let's calculate the moles of each reactant. For Na₃PO₄, we have:
- Volume = 15.48 ml = 0.01548 L
- Concentration = 0.31 M
- Moles of Na₃PO₄ = Volume × Concentration = 0.01548 L × 0.31 M = 0.0047998 mol
For Pb(NO₃)₂:
- Volume = 14.66 ml = 0.01466 L
- Concentration = 0.2 M
- Moles of Pb(NO₃)₂ = Volume × Concentration = 0.01466 L × 0.2 M = 0.002932 mol
According to the reaction, 2 moles of Na₃PO₄ react with 3 moles of Pb(NO₃)₂. To find the limiting reactant, we must compare the mole ratio of the reactants with the stoichiometric ratio from the balanced equation. Using the calculated moles, we find that Pb(NO₃)₂ is the limiting reactant. This means that the amount of Pb₃(PO₄)₂ (lead(II) phosphate) formed will be based on the moles of Pb(NO₃)₂ available.