Problem 8 - 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? \(\mathrm{Al}(\mathrm{OH})_{3}(\mathrm{s})+3 \mathrm{HCl}(\mathrm{aq}) \longrightarrow \mathrm{AlCl}_{3}(\mathrm{aq})+3 \mathrm{H}_{2} \mathrm{O}(\ell)\)

Step-by-Step Solution

Answer

1.052 g of HCl is needed and 0.520 g of water is produced.

Step 1: Determine Molar Masses

First, we calculate the molar mass of each compound involved. \( \mathrm{Al(OH)_3} \) has a molar mass of \( \text{Al (26.98 g/mol)} + 3 \times (\text{O (16.00 g/mol)} + \text{H (1.01 g/mol)}) = 78.00 \text{ g/mol} \). \( \mathrm{HCl} \) has a molar mass of \( \text{H (1.01 g/mol)} + \text{Cl (35.45 g/mol)} = 36.46 \text{ g/mol} \). Water, \( \mathrm{H_2O} \), has a molar mass of \( 2 \times \text{H (1.01 g/mol)} + \text{O (16.00 g/mol)} = 18.02 \text{ g/mol} \).

Step 2: Convert Mass to Moles of Al(OH)_3

Use the molar mass of \( \mathrm{Al(OH)_3} \) to convert the given mass to moles. \( 0.750 \text{ g} \div 78.00 \text{ g/mol} = 0.00962 \text{ mol} \) of \( \mathrm{Al(OH)_3} \).

Step 3: Moles of HCl Required

From the balanced chemical equation, 3 moles of \( \mathrm{HCl} \) are required for every mole of \( \mathrm{Al(OH)_3} \). Calculate the moles of \( \mathrm{HCl} \) needed: \( 0.00962 \text{ mol} \times 3 = 0.02886 \text{ mol} \) of \( \mathrm{HCl} \).

Step 4: Convert Moles of HCl to Mass

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

Step 5: Moles of Water Produced

From the balanced chemical equation, 3 moles of \( \mathrm{H_2O} \) are produced per mole of \( \mathrm{Al(OH)_3} \). Thus, the moles of \( \mathrm{H_2O} \) generated: \( 0.00962 \text{ mol} \times 3 = 0.02886 \text{ mol} \).

Step 6: Convert Moles of Water to Mass

Convert the moles of \( \mathrm{H_2O} \) to grams: \( 0.02886 \text{ mol} \times 18.02 \text{ g/mol} = 0.520 \text{ g} \) of \( \mathrm{H_2O} \).

Key Concepts

Understanding Molar Mass

Molar mass is a fundamental concept in stoichiometry that represents the mass of one mole of a substance. It is expressed in grams per mole (g/mol) and allows us to convert between the mass of a substance and the number of moles. To find the molar mass, we simply add the atomic masses of all the atoms in a molecule, considering their respective proportions. For example:
* Aluminum hydroxide, \( \mathrm{Al(OH)_3} \), has a molar mass calculated as follows: * Al contributes 26.98 g/mol. * Each OH group contributes \(16.00 \text{ g/mol (for O)} + 1.01 \text{ g/mol (for H)} = 17.01 \text{ g/mol}\). * Therefore, \(3 \times 17.01 \text{ g/mol} = 51.03 \text{ g/mol}\) from three OH groups. * Total molar mass = \(26.98 \text{ g/mol} + 51.03 \text{ g/mol} = 78.01 \text{ g/mol}\).Similarly, the molar mass for HCl is 36.46 g/mol, derived from 1.01 g/mol for H and 35.45 g/mol for Cl. Calculating molar mass accurately is crucial for correctly converting between mass and moles, especially in chemical reactions.

Chemical Reactions and Stoichiometry

Chemical reactions involve the conversion of reactants into products, following the conservation of mass principle. To understand a chemical reaction like the one given:\[\mathrm{Al(OH)_3(s) + 3 HCl(aq) \longrightarrow AlCl_3(aq) + 3 H_2O(\ell)}\]we must focus on the balanced chemical equation. This equation indicates the molar relationship between reactants and products:* One mole of \( \mathrm{Al(OH)_3} \) reacts with three moles of \( \mathrm{HCl} \).* This reaction produces one mole of \( \mathrm{AlCl_3} \) and three moles of \( \mathrm{H_2O} \).
The coefficients in a balanced equation reveal the stoichiometric relationships, which are essential for determining how much of each substance will participate in the reaction. Mastering these relationships allows us to calculate the required or produced quantities of a substance, as seen in converting mass in grams to moles and vice versa.

Mass-to-Mole Conversion Techniques

Mass-to-mole conversion is a pivotal stoichiometric technique used to relate the amount of material (in mass) to the amount in moles. This conversion facilitates understanding and performing chemical reaction calculations. The conversion process is typically done through the equation:
\[ \text{number of moles} = \frac{\text{mass in grams}}{\text{molar mass in } \text{g/mol}} \]
For instance, when given 0.750 grams of \( \mathrm{Al(OH)_3} \), and the molar mass is 78.00 g/mol, we calculate moles as follows:
\[ 0.750 \text{ g} \div 78.00 \text{ g/mol} = 0.00962 \text{ mol} \]
Performing such a conversion allows subsequent application of stoichiometry to determine, for example, the amount of \( \mathrm{HCl} \) required for a reaction, or to quantify the products formed. Being precise in these conversions is crucial to achieving accurate and useful results in chemistry lab settings and theoretical calculations.