Question

Determine the volume of sulfuric acid solution needed to prepare \(37.4 \mathrm{~g}\) of aluminum sulfate, \(\mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3}\), by the reaction $$ 2 \mathrm{Al}(s)+3 \mathrm{H}_{2} \mathrm{SO}_{4}(a q) \longrightarrow \mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3}(a q)+3 \mathrm{H}_{2}(g) $$ The sulfuric acid solution, whose density is \(1.104 \mathrm{~g} / \mathrm{mL}\) contains \(15.0 \% \mathrm{H}_{2} \mathrm{SO}_{4}\) by mass. 4.160 Determine the volume of sodium hydroxide solution needed to prepare \(26.2 \mathrm{~g}\) sodium phosphate, \(\mathrm{Na}_{3} \mathrm{PO}_{4}\), by the reaction $$ 3 \mathrm{NaOH}(a q)+\mathrm{H}_{3} \mathrm{PO}_{4}(a q) \longrightarrow \mathrm{Na}_{3} \mathrm{PO}_{4}(a q)+3 \mathrm{H}_{2} \mathrm{O}(l) $$ The sodium hydroxide solution, whose density is \(1.133 \mathrm{~g} /\) \(\mathrm{mL},\) contains \(12.0 \% \mathrm{NaOH}\) by mass.

Short answer

Use 193.55 mL of sulfuric acid solution and 141.2 mL of sodium hydroxide solution.

Step-by-step solution

Calculate moles of Aluminum Sulfate

First, determine the molar mass of aluminum sulfate, \( \mathrm{Al}_2(\mathrm{SO}_4)_3 \), by adding the atomic masses (\(2 \times 27\) for \(\mathrm{Al}\), \(3 \times 32.07\) for \(\mathrm{S}\), and \(12 \times 16\) for \(\mathrm{O}\)). This gives \(342.15\,\mathrm{g/mol}\). Next, calculate moles of \(\mathrm{Al}_2(\mathrm{SO}_4)_3\) by dividing the given mass \(37.4\,\mathrm{g}\) by \(342.15\,\mathrm{g/mol}\), resulting in approximately \(0.109\,\mathrm{mol}\).

Determine moles of Sulfuric Acid required

According to the balanced chemical equation, 3 moles of \(\mathrm{H}_2\mathrm{SO}_4\) are required to produce 1 mole of \(\mathrm{Al}_2(\mathrm{SO}_4)_3\). Therefore, multiply \(0.109\,\mathrm{mol}\) of \(\mathrm{Al}_2(\mathrm{SO}_4)_3\) by \(3\), which equals \(0.327\,\mathrm{mol}\) of \(\mathrm{H}_2\mathrm{SO}_4\) needed.

Calculate mass of Sulfuric Acid

Find the molar mass of \(\mathrm{H}_2\mathrm{SO}_4\), which is \(98.08\,\mathrm{g/mol}\). Then, multiply \(0.327\,\mathrm{mol}\) by \(98.08\,\mathrm{g/mol}\) to find the mass of \(\mathrm{H}_2\mathrm{SO}_4\), equaling approximately \(32.06\,\mathrm{g}\).

Adjust for solution concentration

The sulfuric acid solution is 15% \(\mathrm{H}_2\mathrm{SO}_4\) by mass, meaning 100\,\mathrm{g} of solution contains 15\,\mathrm{g} \(\mathrm{H}_2\mathrm{SO}_4\). Find the total mass of solution needed: \(\frac{32.06\,\mathrm{g}}{0.15}\approx213.73\,\mathrm{g}\).

Calculate volume of Sulfuric Acid solution

Finally, convert solution mass to volume using the solution density \(1.104\,\mathrm{g/mL}\). Thus, \(\frac{213.73\,\mathrm{g}}{1.104\,\mathrm{g/mL}}\approx193.55\,\mathrm{mL}\).

Calculate moles of Sodium Phosphate

First, find the molar mass of \(\mathrm{Na}_3\mathrm{PO}_4\), which is \(3 \times 23\) for \(\mathrm{Na}\), \(31\) for \(\mathrm{P}\), and \(4 \times 16\) for \(\mathrm{O}\). This equals \(163.94\,\mathrm{g/mol}\). Then, calculate moles by dividing mass \(26.2\,\mathrm{g}\) by \(163.94\,\mathrm{g/mol}\), resulting in approximately \(0.160\,\mathrm{mol}\).

Determine moles of Sodium Hydroxide required

The balanced reaction shows that 3 moles of \(\mathrm{NaOH}\) produce 1 mole of \(\mathrm{Na}_3\mathrm{PO}_4\), so use \(0.160\,\mathrm{mol} \times 3 = 0.480\,\mathrm{mol}\) of \(\mathrm{NaOH}\).

Calculate mass of Sodium Hydroxide

With \(0.480\,\mathrm{mol}\) \(\mathrm{NaOH}\) and a molar mass of \(40\,\mathrm{g/mol}\), the mass is \(0.480 \times 40 = 19.2\,\mathrm{g}\).

Adjust for solution concentration

The \(\mathrm{NaOH}\) solution is 12% by mass, so \(19.2\,\mathrm{g}\) represents only 12% of the total mass. Calculate total solution mass with \(\frac{19.2}{0.12} = 160\,\mathrm{g}\).

Calculate volume of Sodium Hydroxide solution

Convert mass of solution to volume using density \(1.133\,\mathrm{g/mL}\). The volume is \(\frac{160\,\mathrm{g}}{1.133\,\mathrm{g/mL}}\approx141.2\,\mathrm{mL}\).

Key concepts

Chemical Reactions

Chemical reactions are processes where substances, called reactants, transform into different substances, known as products. They are vital in chemistry because they show how materials interact and change under certain conditions.
In the given exercise, we examine two distinct chemical reactions. The first reaction involves aluminum reacting with sulfuric acid to produce aluminum sulfate and hydrogen gas:

• 2 Al(s) + 3 H\(_{2}\)SO\(_{4}\)(aq) → Al\(_{2}\)(SO\(_{4}\))\(_{3}\)(aq) + 3 H\(_{2}\)(g)

The second reaction is between sodium hydroxide and phosphoric acid yielding sodium phosphate and water:

• 3 NaOH(aq) + H\(_{3}\)PO\(_{4}\)(aq) → Na\(_{3}\)PO\(_{4}\)(aq) + 3 H\(_{2}\)O(l)

Understanding these equations is crucial because every component in the reaction has a specific role and relationship with others.
Balancing chemical reactions ensures the correct proportion of molecules and the conservation of mass, vital for stoichiometry calculations. It helps to predict the quantities needed and produced, crucial for practical applications in labs and industry.

Molar Mass Calculations

Molar mass calculations help us determine how much of a substance is involved in a chemical reaction. Calculating molar mass involves summing the atomic masses of all atoms in a molecule based on the periodic table.
For instance, aluminum sulfate, Al\(_{2}\)(SO\(_{4}\))\(_{3}\), is made up of

• 2 aluminum atoms (Al), each with an atomic mass of 27 g/mol,
• 3 sulfur atoms (S), each 32.07 g/mol,
• 12 oxygen atoms (O), each 16 g/mol.

Thus, its molar mass is 342.15 g/mol, calculated as \(2 \times 27 + 3 \times 32.07 + 12 \times 16\).
These calculations allow us to convert mass into moles, an essential step for further stoichiometry. For example, with 37.4 g of aluminum sulfate, dividing by its molar mass, 342.15 g/mol, gives approximately 0.109 mol of Al\(_{2}\)(SO\(_{4}\))\(_{3}\).
Such conversions are pivotal because chemical reactions proceed according to the number of moles, not simply mass, adhering to principles of matter conservation during reactions.

Concentration and Solutions

Solutions are mixtures where substances (solutes) dissolve in solvents. Concentration indicates how much solute is present in a given amount of solution. Understanding this allows us to calculate the amount of solution required for reactions, based on its solute content.
In the exercise, sulfuric acid (H\(_{2}\)SO\(_{4}\)) and sodium hydroxide (NaOH) solutions are used, where concentrations are given as a percentage by mass. This percentage describes the mass of solute per 100 g of solution.

• 15% H\(_{2}\)SO\(_{4}\) means 15 g of H\(_{2}\)SO\(_{4}\) per 100 g of solution.
• 12% NaOH means 12 g of NaOH per 100 g of solution.

To know how much of these solutions are needed, convert mass of the solutes into solution volumes via their densities (1.104 g/mL for H\(_{2}\)SO\(_{4}\) and 1.133 g/mL for NaOH).
By dividing solution mass by its density, we convert mass to volume, a necessary step to practically measure liquid reactants for chemical reactions. This understanding is crucial for experimental accuracy and efficiency.