Aluminum ions react with carbonate ions to form an insoluble compound, aluminum carbonate. (a) Write the net ionic equation for this reaction. (b) What is the molarity of a solution of aluminum chloride if \(30.0 \mathrm{~mL}\) is required to react with \(35.5 \mathrm{~mL}\) of \(0.137 \mathrm{M}\) sodium carbonate? (c) How many grams of aluminum carbonate are formed in (b)?

First, we need to balance the equation by making sure that there are equal number of atoms of each element on both sides. The balanced equation is: \(\mathrm{Al^{3+}}+3\mathrm{CO_{3}^{2{-}}}\rightarrow\mathrm{Al{(CO_3)}_3}\) The net ionic equation is the same as the balanced equation since there are no spectator ions present: \(\mathrm{Al^{3+}}+3\mathrm{CO_{3}^{2{-}}}\rightarrow\mathrm{Al{(CO_3)}_3}\)

In this step, we will use the concept of stoichiometry to find the molarity of aluminum chloride. We know the volume and molarity of sodium carbonate solution. It's given as: \(V_{Na_2CO_3}=35.5\,\mathrm{mL}\) \(C_{Na_2CO_3}=0.137\,\mathrm{M}\) We also know that the balanced equation for the reaction between sodium carbonate and aluminum chloride is: \(3\mathrm{Na_2CO_3}+\mathrm{2AlCl_3}\rightarrow\mathrm{6NaCl+2Al(CO_3)_3}\) To determine the molarity of aluminum chloride, we can use the stoichiometric coefficients. From the balanced equation, 3 moles of sodium carbonate reacts with 2 moles of aluminum chloride. So, \(C_{AlCl_3}=\frac{2}{3} \times \frac{V_{Na_2CO_3} \times C_{Na_2CO_3}}{V_{AlCl_3}}\) \(C_{AlCl_3}=\frac{2}{3} \times \frac{35.5\,\mathrm{mL} \times 0.137\,\mathrm{M}}{30.0\,\mathrm{mL}}\) \(C_{AlCl_3}=0.103\,\mathrm{M}\)

Now, we need to determine the mass of aluminium carbonate formed during the reaction. We will use the balanced equation and stoichiometry to calculate the mass. From the balanced equation, we know that 3 moles of sodium carbonate reacts with 2 moles of aluminum carbonate. We have the volume and molarity of sodium carbonate. We can determine the mass of aluminum carbonate formed by the following equation: \(Mass_{Al(CO_3)_3} = 2 \times \frac{Molar\:mass_{Al(CO_3)_3}}{3} \times \mathrm{moles\:of\:Na_2CO_3\,}\, \) \(Mass_{Al(CO_3)_3} = 2\times\frac{114.0\,\mathrm{g/mol}}{3}\times (35.5\,\mathrm{mL} \times 0.137\,\mathrm{M})\) \(Mass_{Al(CO_3)_3} = 3.18\,\mathrm{g}\) So, 3.18 grams of aluminum carbonate are formed.