Problem 2 - Chemistry and Chemical Reactivity

Question

What mass of HCl, in grams, is required to react with \(0.750 \mathrm{g}\) of \(\mathrm{Al}(\mathrm{OH})_{3} ?\) What mass of water, in grams, is produced? $$\mathrm{Al}(\mathrm{OH})_{3}(\mathrm{s})+3 \mathrm{HCl}(\mathrm{aq}) \rightarrow \mathrm{AlCl}_{3}(\mathrm{aq})+3 \mathrm{H}_{2} \mathrm{O}(\ell)$$

Step-by-Step Solution

Answer

1.052 g of HCl and 0.520 g of water.

Step 1: Determine Moles of Al(OH)3

Calculate the number of moles of \( \mathrm{Al(OH)}_{3} \) using its molar mass, which is \( 78.00 \mathrm{g/mol} \). Use the mass-given formula: \[ \text{Moles of \( \mathrm{Al(OH)_{3}} \)} = \frac{0.750 \, \text{g}}{78.00 \, \text{g/mol}} \approx 0.00962 \, \text{mol} \]

Step 2: Use Stoichiometry to Find Moles of HCl Needed

According to the balanced equation, \(1 \) mole of \( \mathrm{Al(OH)_{3}} \) reacts with \(3 \) moles of \( \mathrm{HCl} \). Therefore, multiply the moles of \( \mathrm{Al(OH)_{3}} \) by \(3\): \[ 0.00962 \, \text{mol \( \mathrm{Al(OH)_{3}} \)} \times 3 = 0.02886 \, \text{mol \( \mathrm{HCl} \)} \]

Step 3: Calculate Mass of HCl Required

Convert moles of \( \mathrm{HCl} \) to grams using its molar mass, \( 36.46 \, \text{g/mol} \): \[ \text{Mass of \( \mathrm{HCl} \)} = 0.02886 \, \text{mol} \times 36.46 \, \text{g/mol} = 1.052 \, \text{g} \]

Step 4: Use Stoichiometry to Find Moles of Water Produced

From the balanced equation, \(1 \) mole of \( \mathrm{Al(OH)_{3}} \) produces \(3 \) moles of \( \mathrm{H}_2\mathrm{O} \). Therefore, the moles of water produced are: \[ 0.00962 \, \text{mol \( \mathrm{Al(OH)_{3}} \)} \times 3 = 0.02886 \, \text{mol \(\mathrm{H}_2\mathrm{O} \)} \]

Step 5: Calculate Mass of Water Produced

Convert moles of \( \mathrm{H}_2\mathrm{O} \) to grams using its molar mass, \( 18.02 \, \text{g/mol}\): \[ \text{Mass of \( \mathrm{H}_2\mathrm{O} \)} = 0.02886 \, \text{mol} \times 18.02 \, \text{g/mol} = 0.520 \, \text{g} \]

Key Concepts

Chemical Reactions

Chemical reactions are the processes through which substances, known as reactants, convert into different substances, named products. This transformation involves the rearrangement of atoms and can often be observed visually, like the formation of bubbles or the change of color.
In the given exercise, the chemical reaction involves the reactant aluminum hydroxide, \( \mathrm{Al(OH)_{3}} \), reacting with hydrochloric acid, \( \mathrm{HCl} \).

• Reactants: \( \mathrm{Al(OH)_{3}} \) and \( \mathrm{HCl} \)
• Products: \( \mathrm{AlCl_{3}} \) and water \( \mathrm{H_{2}O} \)

The reaction occurs in an aqueous solution, which is common for reactions involving acids like \( \mathrm{HCl} \). Emphasizing the importance of chemical reactions, they are responsible for numerous natural processes and technological applications, from digestion in our bodies to producing energy in batteries.

Moles and Molar Mass

Understanding moles and molar mass is crucial in stoichiometry, which is the calculation of reactants and products in chemical reactions. A mole is a unit representing \( 6.022 \times 10^{23} \) entities, such as atoms or molecules, similar to how a dozen represents 12 items.
The molar mass of a substance is the mass of one mole of its particles measured in grams per mole (g/mol). In this exercise:

• The molar mass of \( \mathrm{Al(OH)_{3}} \) is \( 78.00 \mathrm{g/mol} \), meaning one mole of it weighs 78.00 grams.
• For \( \mathrm{HCl} \), the molar mass is \( 36.46 \mathrm{g/mol} \), and for \( \mathrm{H_{2}O} \), it is \( 18.02 \mathrm{g/mol} \).

Using these values allows us to convert between mass and moles, an integral part of solving stoichiometry problems. This conversion is essential because chemical equations are usually balanced in terms of moles, not grams.

Balanced Chemical Equations

A balanced chemical equation is fundamental to properly describe a chemical reaction because it respects the law of conservation of mass, meaning mass is neither created nor destroyed.
Each side of the equation must have the same number of each type of atom. In the presented equation:

• The number of aluminum (\( \mathrm{Al} \)) atoms remains 1.
• Hydrochloric acid (\( \mathrm{HCl} \)) provides three chlorine (\( \mathrm{Cl} \)) atoms for every formula unit of aluminum chloride (\( \mathrm{AlCl_{3}} \)), and this balance is maintained.
• The water's hydrogen and oxygen atoms are also perfectly accounted for with three \( \mathrm{H_{2}O} \) molecules compensating for three hydroxide groups from \( \mathrm{Al(OH)_{3}} \).

This equation allows understanding of the proportions in which reactants react and products are formed, aiding in quantitative analysis like calculating precise amounts needed or produced. Balancing equations ensures that stoichiometric calculations yield correct and meaningful results.