A solution of \(105.0 \mathrm{~mL}\) of \(0.300 \mathrm{M} \mathrm{NaOH}\) is mixed with a solution of \(150.0 \mathrm{~mL}\) of \(0.060 \mathrm{MAlCl}_{3} .\) (a) Write the balanced chemical equation for the reaction that occurs. (b) What precipitate forms? (c) What is the limiting reactant? (d) How many grams of this precipitate form? (e) What is the concentration of each ion that remains in solution?

First, let's write the balanced chemical equation for the reaction between Sodium Hydroxide (NaOH) and Aluminum Chloride (AlCl3): \[ \ce{3 NaOH + AlCl3 -> Al(OH)3(s) + 3NaCl} \]

The precipitate formed in this reaction is Aluminum Hydroxide (Al(OH)3), as it is a solid (s).

Now, we will find the limiting reactant by calculating the moles of each reactant and comparing their mole ratio in the balanced equation: Moles of NaOH: \[moles = volume (L) \times molarity \] \[moles_{NaOH} = 0.105 \mathrm{~L} \times 0.300 \mathrm{M} = 0.0315 \mathrm{~mol} \] Moles of AlCl3: \[moles_{AlCl_3} = 0.150 \mathrm{~L} \times 0.060 \mathrm{M} = 0.0090 \mathrm{~mol}\] Now we can compare the moles of reactants with the stoichiometry of the balanced equation. For every mole of AlCl3, we need 3 moles of NaOH. Like this: \[0.0090 \mathrm{~mol~AlCl}_3 \times \frac{3 \mathrm{~mol~NaOH}}{1 \mathrm{~mol~AlCl_{3}}} = 0.0270 \mathrm{~mol~NaOH}\] Since 0.0270 mol of NaOH is required for complete reaction with 0.0090 mol AlCl3, but we have 0.0315 mol of NaOH, there is excess NaOH in the reaction. Therefore, AlCl3 is the limiting reactant.

Now, let's find out how many grams of Aluminum Hydroxide (Al(OH)3) are formed. We use the moles of the limiting reactant (AlCl3) and the balanced chemical equation: \[0.0090 \mathrm{~mol~AlCl}_3 \times \frac{1 \mathrm{~mol~Al(OH)_{3}}}{1 \mathrm{~mol~AlCl_{3}}} \times \frac{78.0 \mathrm{~g}}{1 \mathrm{~mol~Al(OH)_{3}}} = 0.702 \mathrm{~g~Al(OH)_{3}}\] So, 0.702 grams of Aluminum Hydroxide (Al(OH)3) are formed in the reaction.

Finally, we can find the concentration of the remaining ions in the solution. Since AlCl3 is the limiting reactant, we can assume all of it has reacted and no Al3+ or Cl- ions will be left in solution. For Na+ and OH- ions, we only need to consider the excess NaOH in the solution: Excess moles of NaOH: \[0.0315 \mathrm{~mol} - 0.0270 \mathrm{~mol} = 0.0045 \mathrm{~mol}\] As the volume of the reaction mixture is the sum of the volumes of the two initial solutions: \[volume = 0.105 \mathrm{~L} + 0.150 \mathrm{~L} = 0.255 \mathrm{~L}\] Now, let's find the concentration of Na+ and OH- ions in the solution: \[concentration_{Na^+} = concentration_{OH^-} = \frac{0.0045 \mathrm{~mol}}{0.255 \mathrm{~L}} = 0.0176 \mathrm{~M}\] So, the concentrations of Na+ and OH- ions remaining in the solution are both 0.0176 M.