Problem 63 - Chemistry The Central Science In Si Units

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 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 equation is \(\mathrm{Al(OH)}_{3} + 3\mathrm{HCl} \rightarrow \mathrm{AlCl}_{3} + 3\mathrm{H}_{2} \mathrm{O}\). For 0.500 g of Al(OH)₃, 0.701 g of HCl reacts, forming 0.856 g of AlCl₃ and 0.346 g of H₂O. Mass is conserved.

Step 1: Balance the Chemical Equation

The unbalanced equation is:\[ \mathrm{Al(OH)}_{3}(s) + \mathrm{HCl}(aq) \rightarrow \mathrm{AlCl}_{3}(aq) + \mathrm{H}_{2} \mathrm{O}(l) \]First, balance the Al, Cl, and O atoms:1. Aluminum (Al) is already balanced.2. Chlorine (Cl) needs to be balanced. We need 3 HCl to form 1 AlCl₃, so we have: \[ \mathrm{Al(OH)}_{3}(s) + 3\mathrm{HCl}(aq) \rightarrow \mathrm{AlCl}_{3}(aq) + 3\mathrm{H}_{2} \mathrm{O}(l) \]3. Count the oxygens and hydrogens. We have 3 OH groups, which requires 3 HCl molecules producing 3 water molecules (3 H₂O makes 3 O and 6 H atoms needed). The equation is now balanced.The balanced equation is:\[ \mathrm{Al(OH)}_{3}(s) + 3\mathrm{HCl}(aq) \rightarrow \mathrm{AlCl}_{3}(aq) + 3\mathrm{H}_2\mathrm{O}(l) \]

Step 2: Calculate Moles of Al(OH)₃

To find moles of \(\mathrm{Al(OH)}_{3}\), use its molar mass:- Molar mass of Al(OH)₃: Al = 27.0 g/mol, O = 16.0 g/mol, H = 1.0 g/mol \[ \text{Molar mass of } \mathrm{Al(OH)}_{3} = 27.0 + 3(16.0+1.0) = 78.0 \, \mathrm{g/mol} \]Calculate moles:\[ \text{Moles of } \mathrm{Al(OH)}_{3} = \frac{0.500 \, \mathrm{g}}{78.0 \, \mathrm{g/mol}} \approx 0.00641 \, \mathrm{mol} \]

Step 3: Calculate Grams of HCl Required

Using the balanced equation, the mole ratio between \(\mathrm{Al(OH)}_{3}\) and \(\mathrm{HCl}\) is 1:3.Calculate moles of HCl:\[ \text{Moles of } \mathrm{HCl} = 3 \times 0.00641 \, \mathrm{mol} = 0.0192 \, \mathrm{mol} \]The molar mass of HCl is 36.5 g/mol.Calculate grams of HCl:\[ \text{Mass of } \mathrm{HCl} = 0.0192 \, \mathrm{mol} \times 36.5 \, \mathrm{g/mol} \approx 0.701 \, \mathrm{g} \]

Step 4: Calculate Grams of Products (AlCl₃ and H₂O) Formed

Using the balanced equation, the mole ratios are:- \(\mathrm{Al(OH)}_{3}\) to \(\mathrm{AlCl}_{3} = 1:1\)- \(\mathrm{Al(OH)}_{3}\) to \(\mathrm{H}_{2} \mathrm{O} = 1:3\)Calculate moles and mass of each product:**AlCl₃:** Moles: \(0.00641 \, \mathrm{mol}\) Molar mass: Al = 27.0, Cl = 35.5, \[ \text{Molar mass of } \mathrm{AlCl}_{3} = 27.0 + 3(35.5) = 133.5 \, \mathrm{g/mol} \] \[ \text{Mass of } \mathrm{AlCl}_{3} = 0.00641 \, \mathrm{mol} \times 133.5 \, \mathrm{g/mol} \approx 0.856 \, \mathrm{g} \]**H₂O:** Moles: \(0.00641 \, \mathrm{mol} \times 3 = 0.0192 \, \mathrm{mol}\) Molar mass of H₂O = 18.0 g/mol \[ \text{Mass of } \mathrm{H}_{2} \mathrm{O} = 0.0192 \, \mathrm{mol} \times 18.0 \, \mathrm{g/mol} \approx 0.346 \, \mathrm{g} \]

Step 5: Verify Conservation of Mass

Confirm the mass balance:Total mass of reactants = Mass of \(\mathrm{Al(OH)}_{3}\) + Mass of \(\mathrm{HCl}\):\[ 0.500 \, \mathrm{g} + 0.701 \, \mathrm{g} = 1.201 \, \mathrm{g} \]Total mass of products = Mass of \(\mathrm{AlCl}_{3}\) + Mass of \(\mathrm{H}_{2} \mathrm{O}\):\[ 0.856 \, \mathrm{g} + 0.346 \, \mathrm{g} = 1.202 \, \mathrm{g} \]The small discrepancy is due to rounding in calculations, but mass is nearly conserved, demonstrating the law of conservation of mass.

Key Concepts

Chemical Reactions

Chemical reactions are processes where substances, known as reactants, convert into new substances, called products. Understanding this concept is crucial in chemistry as it is the foundation for predicting how substances interact and change under different conditions. An antacid, such as aluminum hydroxide (\(\mathrm{Al(OH)}_3\)), is a perfect example of a chemical reaction in action. When it meets hydrochloric acid (\(\mathrm{HCl}\)) in the stomach, a reaction occurs forming aluminum chloride (\(\mathrm{AlCl}_3\)) and water (\(\mathrm{H}_2\mathrm{O}\)).
In the case of the antacid reacting with stomach acid, the balanced chemical equation tells the whole story. Initially unbalanced with unequal amounts of chlorine and hydrogen atoms, through analysis, we balance the equation to ensure each element has the same number of atoms on the reactant and product sides. Here's the balanced equation:
\[ \mathrm{Al(OH)}_{3}(s) + 3\mathrm{HCl}(aq) \rightarrow \mathrm{AlCl}_{3}(aq) + 3\mathrm{H}_2\mathrm{O}(l) \]
This shows each component—aluminum, chlorine, and oxygen—is conserved, with no atoms lost during the reaction.

Law of Conservation of Mass

The law of conservation of mass is a fundamental principle in chemistry, stating that mass cannot be created or destroyed in a chemical reaction. This means that the total mass of reactants must equal the total mass of products. This concept assures that chemical reactions are predictable and behave in a consistent manner, something we observed in the antacid reaction.
When calculating the mass of substances before and after the reaction of \(\mathrm{Al(OH)}_3\) with \(\mathrm{HCl}\), we found:

• Total mass of reactants (\(\mathrm{Al(OH)}_3\) and \(\mathrm{HCl}\)): \(0.500 \ \,\mathrm{g} + 0.701 \,\mathrm{g} = 1.201 \,\mathrm{g}\)
• Total mass of products (\(\mathrm{AlCl}_3\) and \(\mathrm{H_2O}\)): \(0.856 \,\mathrm{g} + 0.346 \,\mathrm{g} = 1.202 \,\mathrm{g}\)

The slight difference in masses is attributed to rounding errors in calculations. However, in theory, these should be equal, demonstrating that this law holds true within these stoichiometric calculations.
This concept is essential, ensuring that equations are appropriately balanced to reflect accurately the behavior of chemical reactions.

Mole Calculations

Mole calculations are a cornerstone of stoichiometry, crucial for converting between masses of substances and amounts in moles. This allows one to quantitatively describe a reaction. The mole is a unit that bridges the atomic scale with the macroscopic scale we can measure.
For example, in the reaction of \(\mathrm{Al(OH)}_3\) with \(\mathrm{HCl}\), you would begin by calculating the number of moles of \(\mathrm{Al(OH)}_3\) given.
Using its molar mass of \(78.0 \,\mathrm{g/mol}\), we find:
\[ \text{Moles of } \mathrm{Al(OH)}_3 = \frac{0.500 \,\mathrm{g}}{78.0 \,\mathrm{g/mol}} \approx 0.00641 \,\mathrm{mol}\]
With the balanced chemical equation, the stoichiometric ratios provide direct relationships between moles of reactants and products. For example, 1 mole of \(\mathrm{Al(OH)}_3\) will react with 3 moles of \(\mathrm{HCl}\) to produce 1 mole of \(\mathrm{AlCl}_3\) and 3 moles of \(\mathrm{H}_2\mathrm{O}\).
Calculating the moles and mass of the reactants and products helps validate the stoichiometry of the reaction. It ensures that the amounts are consistent with the balanced chemical equation, demonstrating the reaction is proceeding as expected.