Key Concept: Complex Reaction Analysis, Real-world Application
b) 98.09 g
[Solution Description] The balanced chemical equation for the reaction is:
$\text{CaCO}_3 + \text{H}_2\text{SO}_4 \rightarrow \text{CaSO}_4 + \text{CO}_2 + \text{H}_2\text{O}$
The molar mass of $\text{CaCO}_3$ is $40.08 + 12.01 + 3 \times 16.00 = 100.09 \, \text{g/mol}$.
Hence, moles of $\text{CaCO}_3$ used are:$\frac{100 \, \text{g}}{100.09 \, \text{g/mol}} \approx 1.00 \, \text{mol}$
From the stoichiometry of the reaction (1:1), moles of $\text{H}_2\text{SO}_4$ required are also $1.00 \, \text{mol}$.
The molar mass of $\text{H}_2\text{SO}_4$ is $2 \times 1.01 + 32.07 + 4 \times 16.00 = 98.09 \, \text{g/mol}$.
Therefore, the mass of $\text{H}_2\text{SO}_4$ neutralized is:$1.00 \, \text{mol} \times 98.09 \, \text{g/mol} = 98.09 \, \text{g}$
Notes:
- Complex Reactions: Reactions involving multiple steps, where products from one stage act as reactants for another.
- Example: Neutralization reaction (acid + base → salt + water), which can involve complex interactions, especially in buffer solutions.
- Real-World Applications:
- Antacid Tablets: Contain bases to neutralize stomach acid.
- Soil Treatment: Lime (a base) is used to neutralize acidic soils for better crop growth.
- Titration: Used in laboratories to determine the concentration of an acid or base in a solution.
- Biological Processes: Blood pH regulation involves complex acid-base reactions to maintain homeostasis.
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b) 98.09 g
[Solution Description] The balanced chemical equation for the reaction is:
$\text{CaCO}_3 + \text{H}_2\text{SO}_4 \rightarrow \text{CaSO}_4 + \text{CO}_2 + \text{H}_2\text{O}$
The molar mass of $\text{CaCO}_3$ is $40.08 + 12.01 + 3 \times 16.00 = 100.09 \, \text{g/mol}$.
Hence, moles of $\text{CaCO}_3$ used are:$\frac{100 \, \text{g}}{100.09 \, \text{g/mol}} \approx 1.00 \, \text{mol}$
From the stoichiometry of the reaction (1:1), moles of $\text{H}_2\text{SO}_4$ required are also $1.00 \, \text{mol}$.
The molar mass of $\text{H}_2\text{SO}_4$ is $2 \times 1.01 + 32.07 + 4 \times 16.00 = 98.09 \, \text{g/mol}$.
Therefore, the mass of $\text{H}_2\text{SO}_4$ neutralized is:$1.00 \, \text{mol} \times 98.09 \, \text{g/mol} = 98.09 \, \text{g}$