Question

Upon heating \(125 \mathrm{~g} \mathrm{AlCl}_{3} \cdot 6 \mathrm{H}_{2} \mathrm{O}\) : (a) how many grams of water can be obtained? (b) how many grams of anhydrous compound can be obtained?

Short answer

61.50 grams of water and 63.50 grams of anhydrous \(\text{AlCl}_3\).

Step-by-step solution

- Determine molar masses

Calculate the molar masses of \(\text{AlCl}_3 \cdot 6 \text{H}_2 \text{O}\), \(\text{H}_2 \text{O}\), and \(\text{AlCl}_3\). For \(\text{AlCl}_3 \cdot 6 \text{H}_2 \text{O}\), molar mass = 133.35 \text{g/mol} + 6 \times 18.02 \text{g/mol} = 219.47 \text{g/mol}. For \(\text{H}_2 \text{O}\), molar mass = 18.02 \text{g/mol}. For \(\text{AlCl}_3\), molar mass = 133.35 \text{g/mol}.

- Calculate moles of hydrated compound

Use the given mass to find the moles of \(\text{AlCl}_3 \cdot 6 \text{H}_2 \text{O}\). Using the formula \[ \text{moles} = \frac{\text{mass}}{\text{molar mass}} \] we have \(\text{moles} = \frac{125 \text{g}}{219.47 \text{g/mol}} \approx 0.569 \text{moles}\).

- Calculate mass of water

Since each mole of \(\text{AlCl}_3 \cdot 6 \text{H}_2 \text{O}\) yields 6 moles of \(\text{H}_2 \text{O}\), the number of moles of water is \(0.569 \times 6 \approx 3.414 \text{moles}\). Then, the mass of water is \[ \text{mass} = \text{moles} \times \text{molar mass} \] which calculates to \(3.414 \times 18.02 \approx 61.50 \text{g}\).

- Calculate mass of anhydrous compound

Find the mass of the anhydrous compound (\text{AlCl}_3) by subtracting the mass of water from the original mass of the hydrated compound: \[ 125 \text{g} - 61.50 \text{g} = 63.50 \text{g} \].

Key concepts

Molar Mass Determination

Molar mass is a crucial concept in stoichiometry and chemistry. It represents the mass of one mole of a substance, typically expressed in grams per mole (g/mol). To determine the molar mass, you sum the atomic masses of all atoms in a molecule based on the periodic table values.
For example, the molar mass of \(\text{AlCl}_3 \cdot 6 \text{H}_2 \text{O}\) includes the atomic masses of aluminum (Al), chlorine (Cl), and water (H2O). With AlCl3, you have 1 Al (26.98 g/mol) and 3 Cl (3 \times 35.45 g/mol), giving a subtotal of 133.35 g/mol. The six water molecules contribute 6 \times 18.02 g/mol, so combining them results in a total molar mass of 219.47 g/mol for the hydrated compound.
This calculated molar mass is used to convert a known mass to moles or vice versa, enabling further quantitative analyses in different chemical reactions.

Chemical Reactions

Chemical reactions involve the transformation of reactants into products through breaking and forming chemical bonds.
For stoichiometric calculations, understanding the balanced equation and the ratios in which substances react is important. Consider the compound \(\text{AlCl}_3 \cdot 6 \text{H}_2 \text{O}\) being heated. In this reaction, the hydrate decomposes, releasing water and leaving behind an anhydrous compound, AlCl3.
By using stoichiometry, we can predict the outcomes of such reactions. Starting with 125 g of \(\text{AlCl}_3 \cdot 6 \text{H}_2 \text{O}\), we convert this mass into moles by dividing it by its molar mass (219.47 g/mol), resulting in approximately 0.569 moles of the hydrate. For each mole of hydrate, 6 moles of water (\textrm{H}_2{O}) are produced. Finally, by converting the moles of water into grams (0.569 moles \times 18.02 g/mol), we determine that around 61.50 g of \textrm{H}_2{O} is produced.

Hydrated Compounds

Hydrated compounds contain water molecules within their crystalline structures. These water molecules, known as waters of hydration, are integral parts of the compound but can be removed through heating or other chemical processes.
In the exercise, \(\text{AlCl}_3 \cdot 6 \text{H}_2 \text{O}\) is a sample of a hydrated compound where six water molecules are chemically integrated into its structure. When heated, it releases these water molecules, transforming into anhydrous \textrm{AlCl}_3.
The notation \(\text{AlCl}_3 \cdot 6 \text{H}_2 \text{O}\) precisely indicates one mole of AlCl3 associated with six moles of water. To find how many grams of water can be obtained following heating, we use the mole concept and molar masses. This involves converting the given mass of the compound to moles and applying the stoichiometric ratio between the hydrated and anhydrous forms to find the amount of water released.