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

1.051 g of HCl is required, and 0.519 g of water is produced.

Step 1: Write the Balanced Chemical Equation

The reaction given is already balanced: \[ \mathrm{Al(OH)}_3(\mathrm{s}) + 3\, \mathrm{HCl}(\mathrm{aq}) \rightarrow \mathrm{AlCl}_3(\mathrm{aq}) + 3\, \mathrm{H_2O}(\ell) \] This equation shows that 1 mole of \( \mathrm{Al(OH)}_3 \) reacts with 3 moles of \( \mathrm{HCl} \) to produce 3 moles of \( \mathrm{H_2O} \).

Step 2: Calculate the Molar Masses

First, calculate the molar masses of \( \mathrm{Al(OH)}_3 \), \( \mathrm{HCl} \), and \( \mathrm{H_2O} \):- \( \mathrm{Al(OH)}_3: \) - Aluminum (Al): 26.98 g/mol - Oxygen (O): 16.00 g/mol - Hydrogen (H): 1.01 g/mol - Total: \( 26.98 + (3 \times 16.00) + (3 \times 1.01) = 78.01 \ \mathrm{g/mol} \)- \( \mathrm{HCl}: \) - Hydrogen (H): 1.01 g/mol - Chlorine (Cl): 35.45 g/mol - Total: \( 1.01 + 35.45 = 36.46 \ \mathrm{g/mol} \)- \( \mathrm{H_2O}: \) - Hydrogen (H): 1.01 g/mol \( \times 2\) - Oxygen (O): 16.00 g/mol - Total: \( (2 \times 1.01) + 16.00 = 18.02 \ \mathrm{g/mol} \)

Step 3: Determine Moles of Al(OH)3

To find the moles of \( \mathrm{Al(OH)}_3 \), use its mass and molar mass:\[ \text{Moles of } \mathrm{Al(OH)}_3 = \frac{0.750\, \mathrm{g}}{78.01 \ \mathrm{g/mol}} = 0.00961 \text{ moles} \]

Step 4: Calculate Moles of HCl Required

Use the stoichiometry from the balanced equation: 1 mole of \( \mathrm{Al(OH)}_3 \) requires 3 moles of \( \mathrm{HCl} \):\[ \text{Moles of } \mathrm{HCl} = 3 \times 0.00961 = 0.02883 \text{ moles} \]

Step 5: Convert Moles of HCl to Mass

Convert the moles of \( \mathrm{HCl} \) to mass:\[ \text{Mass of } \mathrm{HCl} = 0.02883 \text{ moles} \times 36.46 \ \mathrm{g/mol} = 1.051 \ \mathrm{g} \]

Step 6: Calculate Moles of Water Produced

Since 1 mole of \( \mathrm{Al(OH)}_3 \) produces 3 moles of \( \mathrm{H_2O} \), the moles of \( \mathrm{H_2O} \) produced is:\[ \text{Moles of } \mathrm{H_2O} = 3 \times 0.00961 = 0.02883 \text{ moles} \]

Step 7: Convert Moles of Water to Mass

Convert the moles of \( \mathrm{H_2O} \) to mass:\[ \text{Mass of } \mathrm{H_2O} = 0.02883 \text{ moles} \times 18.02 \ \mathrm{g/mol} = 0.519 \ \mathrm{g} \]

Key Concepts

Balanced Chemical Equation

A balanced chemical equation is vital for stoichiometric calculations. It shows the relationship between reactants and products in a chemical reaction. Here, the balanced equation is \( \mathrm{Al(OH)}_3(\mathrm{s}) + 3 \mathrm{HCl}(\mathrm{aq}) \rightarrow \mathrm{AlCl}_3(\mathrm{aq}) + 3 \mathrm{H_2O}(\ell) \).
This equation tells us the ratio in which the substances react and are produced: 1 mole of \( \mathrm{Al(OH)}_3 \) reacts with 3 moles of \( \mathrm{HCl} \) to produce 1 mole of \( \mathrm{AlCl}_3 \) and 3 moles of \( \mathrm{H_2O} \).
Balancing equations ensures that the same number of each type of atom is on both sides of the equation. This reflects the Law of Conservation of Mass, meaning mass is neither created nor destroyed during a chemical reaction.

Molar Mass Calculation

Molar mass is the mass of one mole of a substance, usually measured in grams per mole (g/mol). To calculate molar mass, sum the atomic masses of all atoms in a molecule using values from the periodic table.
For example, the molar mass of \( \mathrm{Al(OH)}_3 \) is calculated as follows:

• Aluminum (Al): 26.98 g/mol
• Oxygen (O): 16.00 g/mol (3 O atoms mean \( 3 \times 16.00 = 48.00 \))
• Hydrogen (H): 1.01 g/mol (3 H atoms mean \( 3 \times 1.01 = 3.03 \))

Total molar mass of \( \mathrm{Al(OH)}_3 \) is 26.98 + 48.00 + 3.03 = 78.01 g/mol.
Similarly, calculate for other compounds like \( \mathrm{HCl} \) and \( \mathrm{H_2O} \):

• \( \mathrm{HCl} \) is 36.46 g/mol
• \( \mathrm{H_2O} \) is 18.02 g/mol

Molar mass is crucial for converting between mass and moles.

Mole Concept

The mole is a basic unit in chemistry that measures the amount of substance. One mole contains Avogadro's number of entities, approximately \( 6.022 \times 10^{23} \).
To determine the number of moles from a given mass, use the formula:
\[\text{Moles} = \frac{\text{Mass (g)}}{\text{Molar Mass (g/mol)}}\]
For instance, with 0.750 g of \( \mathrm{Al(OH)}_3 \) and its molar mass (78.01 g/mol), the calculation is:
\[\text{Moles of } \mathrm{Al(OH)}_3 = \frac{0.750}{78.01} \approx 0.00961 \text{ moles}\]
This concept allows us to link macroscopic measurements to microscopic atoms and molecules.

Stoichiometric Calculations

Stoichiometry involves using balanced chemical equations to calculate reactants and products. It helps in determining how much reactant is needed or how much product will form.
Start with the moles of a known substance, such as \( \mathrm{Al(OH)}_3 \), calculated from mass. Use the stoichiometric ratios from the balanced equation to find the moles of other substances.
For example, from the equation:

• 1 mole of \( \mathrm{Al(OH)}_3 \) reacts with 3 moles of \( \mathrm{HCl} \).
• So, if you have 0.00961 moles of \( \mathrm{Al(OH)}_3 \), it needs \( 3 \times 0.00961 = 0.02883 \) moles of \( \mathrm{HCl} \).

Finally, convert moles back to mass using molar mass,

• e.g., mass of \( \mathrm{HCl} = 0.02883 \times 36.46 = 1.051 \) g
• and mass of \( \mathrm{H_2O} = 0.02883 \times 18.02 = 0.519 \) g

Remember, stoichiometry provides a practical way to predict chemical reactions based on balanced equations.