Problem 63 - Chemistry: The Central Science

Question

Several brands of antacids use \(\mathrm{Al}(\mathrm{OH})_{3}\) to react with stomach acid, which contains primarily HCl: $$ \mathrm{Al}(\mathrm{OH})_{3}(s)+\mathrm{HCl}(a q) \longrightarrow \mathrm{AlCl}_{3}(a q)+\mathrm{H}_{2} \mathrm{O}(l) $$ (a) Balance this equation. (b) Calculate the number of grams of \(\mathrm{HCl}\) that can react with \(0.500 \mathrm{~g}\) of \(\mathrm{Al}(\mathrm{OH})_{3}\) (c) Calculate the number of grams of \(\mathrm{AlCl}_{3}\) and the number of grams of \(\mathrm{H}_{2} \mathrm{O}\) formed when \(0.500 \mathrm{~g}\) of \(\mathrm{Al}(\mathrm{OH})_{3}\) reacts. (d) Show that your calculations in parts (b) and (c) are consistent with the law of conservation of mass.

Step-by-Step Solution

Answer

The balanced chemical equation is: \(\mathrm{Al}(\mathrm{OH})_{3}(s)+3\mathrm{HCl}(a q) \longrightarrow \mathrm{AlCl}_{3}(a q)+3\mathrm{H}_{2} \mathrm{O}(l)\). When 0.500 g of \(\mathrm{Al}(\mathrm{OH})_{3}\) reacts, 0.701 g of HCl is required, and 0.854 g of \(\mathrm{AlCl}_{3}\) and 0.346 g of \(\mathrm{H}_{2} \mathrm{O}\) are formed. The conservation of mass is verified as the mass of reactants (1.201 g) is almost equal to the mass of products (1.200 g).

Step 1: Balancing the chemical equation

Place correct coefficients in front of the reactants and products until the atoms are balanced on both sides. The balanced equation is: $$ \mathrm{Al}(\mathrm{OH})_{3}(s)+3\mathrm{HCl}(a q) \longrightarrow \mathrm{AlCl}_{3}(a q)+3\mathrm{H}_{2} \mathrm{O}(l) $$

Step 2: Calculate the moles of \(\mathrm{Al}(\mathrm{OH})_{3}\)

Using the molar mass of \(\mathrm{Al}(\mathrm{OH})_{3}\), convert grams to moles: Moles of \(\mathrm{Al}(\mathrm{OH})_{3}\) = \(\frac{0.500 \mathrm{~g}}{78.0 \mathrm{~g/mol}}\) = 0.00641 mol

Step 3: Calculate moles of HCl reacted and moles of products formed

Using the balanced equation, we find the stoichiometric relationship between reactants and products: $$ \frac{\text{moles of HCl}}{\text{moles of }\mathrm{Al}(\mathrm{OH})_{3}}=\frac{3}{1} \\ \text{moles of HCl} = 3(0.00641 \text{ mol}) $$ For the products, Moles of \(\mathrm{AlCl}_{3}\) = moles of \(\mathrm{Al}(\mathrm{OH})_{3}\) = 0.00641 mol \\ Moles of \(\mathrm{H}_{2} \mathrm{O}\) = 3 × moles of \(\mathrm{Al}(\mathrm{OH})_{3}\) = 3(0.00641) mol

Step 4: Calculate grams of HCl, \(\mathrm{AlCl}_{3}\), and \(\mathrm{H}_{2} \mathrm{O}\)

Using the molar mass, convert moles to grams: Grams of HCl = 0.01923 mol × \(36.5 \mathrm{~g/mol}\) = 0.701 g \\ Grams of \(\mathrm{AlCl}_{3}\) = 0.00641 mol × \(133.33 \mathrm{~g/mol}\) = 0.854 g \\ Grams of \(\mathrm{H}_{2} \mathrm{O}\) = 0.01923 mol × \(18.0 \mathrm{~g/mol}\) = 0.346 g

Step 5: Verify the law of conservation of mass

According to the law of conservation of mass, the total mass of reactants must equal the total mass of products. Let's check if that holds in our case: Mass of reactants = Mass of \(\mathrm{Al}(\mathrm{OH})_{3}\) + Mass of HCl = \(0.500\mathrm{-} \mathrm{g}+0.701 \mathrm{~g}\) = 1.201 g \\ Mass of products = Mass of \(\mathrm{AlCl}_{3}\) + Mass of \(\mathrm{H}_{2} \mathrm{O}\) = \(0.854 \mathrm{~g} + 0.346 \mathrm{~g}\) = 1.200 g Since the difference between mass of reactants and mass of products is minimal (1.201 g - 1.200 g = 0.001 g), our calculations are consistent with the law of conservation of mass.

Key Concepts

Chemical Reactions

Chemical reactions are processes where substances, known as reactants, transform into different substances called products. Each reaction follows a specific pathway and can be depicted through balanced chemical equations. An equation must have the same number of each type of atom on both sides, reflecting the principle known as the conservation of atoms. For example, in our reaction, aluminum hydroxide (\(\mathrm{Al}(\mathrm{OH})_{3}\)) reacts with hydrochloric acid (\(\mathrm{HCl}\)) to produce aluminum chloride (\(\mathrm{AlCl}_{3}\)) and water (\(\mathrm{H}_{2}\mathrm{O}\)). Balancing the equation involves adjusting coefficients to satisfy this atomic balance.

• Identifying reactants and products: Read the formula to know which substances enter the reaction and those which are formed.
• Balancing the equation: Adjust coefficients in front of compounds to equal the number of each atom in reactants and products. For instance, 3 \(\mathrm{HCl}\) molecules are needed to react with one \(\mathrm{Al}(\mathrm{OH})_{3}\) due to stoichiometry, indicating a 3:1 ratio.

Balancing chemical reactions ensures an accurate depiction of how reactants convert to products.

Molar Mass Calculations

Molar mass calculations help in converting between grams and moles, which is crucial for quantifying the substances in a chemical reaction. The molar mass is the weight of one mole of a compound, usually expressed in grams per mole (g/mol). It's derived from the atomic masses found on the periodic table. For our example:

• Molar mass of \(\mathrm{Al}(\mathrm{OH})_{3}\): To find this, add the atomic masses of aluminum (27 g/mol), oxygen (3 x 16 g/mol), and hydrogen (3 x 1 g/mol), totaling 78 g/mol.
• Converting grams to moles: Use the formula \(\text{Moles} = \frac{\text{grams}}{\text{molar mass}}\). For 0.500 g of \(\mathrm{Al}(\mathrm{OH})_{3}\), divide by 78 g/mol to get approximately 0.00641 moles.

Molar mass calculations help you determine how much of a reactant is needed or how much product will form.

Conservation of Mass

The law of conservation of mass states that mass cannot be created or destroyed in a chemical reaction. This principle ensures that the total mass of reactants equals the total mass of products. For this exercise, the masses of reactants and products are calculated to confirm this law:

• Mass of reactants: Total the masses of \(\mathrm{Al}(\mathrm{OH})_{3}\) and reacted \(\mathrm{HCl}\). For 0.500 g of \(\mathrm{Al}(\mathrm{OH})_{3}\) and 0.701 g of \(\mathrm{HCl}\), the total is 1.201 g.
• Mass of products: Calculate the masses of \(\mathrm{AlCl}_{3}\) and \(\mathrm{H}_{2}\mathrm{O}\). With calculated masses of 0.854 g and 0.346 g, respectively, the total product mass closely equals the reactant mass; any small discrepancy is typically due to rounding in calculations.

Understanding and verifying the conservation of mass in reactions is fundamental for predicting outcomes and balancing equations effectively.