Aluminum sulfate, known as cake alum, has a wide range of uses, from dyeing leather and cloth to purifying sewage. In aqueous solution, it reacts with base to form a white precipitate. (a) Write balanced total and net ionic equations for its reaction with aqueous \(\mathrm{NaOH}\). (b) What mass of precipitate forms when \(185.5 \mathrm{~mL}\) of \(0.533 \mathrm{M} \mathrm{NaOH}\) is added to \(627 \mathrm{~mL}\) of a solution that contains \(15.8 \mathrm{~g}\) of aluminum sulfate per liter?

When aluminum sulfate \(Al_2(SO_4)_3\) reacts with sodium hydroxide \(NaOH\), we can describe the reaction using a balanced molecular equation. A molecular equation provides a complete representation of the chemicals involved and the products formed. Here, aluminum sulfate reacts in aqueous form with sodium hydroxide to form aluminum hydroxide \(Al(OH)_3\), a white precipitate, and sodium sulfate \(Na_2SO_4\), which remains dissolved in water.

The balanced molecular equation for this reaction is as follows: \[ Al_2(SO_4)_3 (aq) + 6 NaOH (aq) \rightarrow 2 Al(OH)_3 (s) + 3 Na_2SO_4 (aq) \]

This equation tells us that one mole of \(Al_2(SO_4)_3\) reacts with six moles of \(NaOH\) to produce two moles of \(Al(OH)_3\) and three moles of \(Na_2SO_4\). It provides a clear, full picture of the reactants converting into products.

To better understand the actual chemical changes occurring in the reaction, we use the net ionic equation. The net ionic equation removes the spectator ions, which are ions that do not participate in the actual precipitation reaction. This focuses only on the ions that form the precipitate. First, we write the full ionic equation by breaking down the aqueous compounds into their ions:

\[ 2 Al^{3+} (aq) + 3 SO_4^{2-} (aq) + 6 Na^+ (aq) + 6 OH^- (aq) \rightarrow 2 Al(OH)_3 (s) + 6 Na^+ (aq) + 3 SO_4^{2-} (aq) \]

In this equation, you can see that \(Na^+\) and \(SO_4^{2-}\) ions appear on both sides. These are the spectator ions and can be omitted to simplify the equation:

\[ 2 Al^{3+} (aq) + 6 OH^- (aq) \rightarrow 2 Al(OH)_3 (s) \]

The net ionic equation succinctly shows that aluminum ions \(Al^{3+}\) react with hydroxide ions \(OH^-\) to form solid aluminum hydroxide \(Al(OH)_3\).

In a chemical reaction, the limiting reactant is the substance that is entirely consumed first, limiting the amount of product formed. To identify the limiting reactant between aluminum sulfate and sodium hydroxide, we must calculate the moles of each reactant using the given molarity and volume.

First, calculate the moles of \(NaOH\):

\[ 0.533 \text{ M} \times 0.1855 \text{ L} = 0.09886 \text{ moles NaOH} \]

Next, calculate the moles of \(Al_2(SO_4)_3\):

The solution contains 15.8 g/L of \(Al_2(SO_4)_3\). Given volume = 0.627 L, convert grams to moles:

\[ \frac{15.8 g}{342.15 g/mol} \times 0.627 L = 0.02895 \text{ moles Al}_2(SO_4)_3 \]

The balanced molecular equation requires 1 mole of \(Al_2(SO_4)_3\) to react with 6 moles of \(NaOH\). Let's find the reactant that produces the lesser amount:

\[ \text{Needed moles of Al}_2(SO_4)_3 = \frac{0.09886 \text{ moles NaOH}}{6} = 0.01648 \text{ moles Al}_2(SO_4)_3 \]

Since 0.02895 moles of \(Al_2(SO_4)_3\) are available but only 0.01648 moles are needed, \(NaOH\) is the limiting reactant. The amount of \(Al(OH)_3\) precipitate formed is based on the limiting reactant moles:

\[ \frac{0.09886 \text{ moles NaOH}}{3} = 0.03295 \text{ moles Al(OH)}_3 \]

Multiply this by the molar mass of \(Al(OH)_3\) (78 g/mol):

\[ 0.03295 \text{ moles} \times 78 \text{ g/mol} = 2.57 \text{ g} \]

The precipitate mass formed is 2.57 grams, dictated by the limiting reactant.