Problem 90 - Chemistry The Central Science In Si Units

Question

A \(1.248-g\) sample of limestone rock is pulverized and then treated with \(30.00 \mathrm{~mL}\) of \(1.035 \mathrm{M}\) HCl solution. The excess acid then requires \(11.56 \mathrm{~mL}\) of \(1.010 \mathrm{M} \mathrm{NaOH}\) for neutralization. Calculate the percentage by mass of calcium carbonate in the rock, assuming that it is the only substance reacting with the HCl solution.

Step-by-Step Solution

Answer

The percentage by mass of calcium carbonate in the rock is approximately 77.56%.

Step 1: Determine Moles of Initial HCl

The initial moles of HCl present in the solution can be calculated using its molarity and volume. Using the formula:\[ \text{moles of HCl initial} = Molarity_{HCl} \times Volume_{HCl} = 1.035 \, \text{M} \times 0.03000 \, \text{L} \]Calculating this gives:\[ \text{moles of HCl initial} = 0.03105 \text{ mol} \]

Step 2: Determine Moles of NaOH Used for Neutralization

The excess HCl is neutralized by NaOH. Calculate the moles of NaOH using its molarity and volume:\[ \text{moles of NaOH} = Molarity_{NaOH} \times Volume_{NaOH} = 1.010 \, \text{M} \times 0.01156 \, \text{L} \]Calculating this gives:\[ \text{moles of NaOH} = 0.0116806 \text{ mol} \]

Step 3: Calculate Moles of Excess HCl

The moles of excess HCl is equal to the moles of NaOH since NaOH completely neutralizes the excess HCl. Therefore:\[ \text{moles of HCl excess} = 0.0116806 \text{ mol} \]

Step 4: Calculate Moles of HCl Reacted with CaCO₃

Subtract the moles of excess HCl from the initial moles of HCl to find out how much reacted with CaCO₃:\[ \text{moles of HCl reacted} = \text{moles of HCl initial} - \text{moles of HCl excess} = 0.03105 \text{ mol} - 0.0116806 \text{ mol} \]This calculates to:\[ \text{moles of HCl reacted} = 0.0193694 \text{ mol} \]

Step 5: Determine Moles of CaCO₃ Reacted

Calcium carbonate reacts with HCl in a 1:2 molar ratio, as per the chemical equation:\[ \text{CaCO}_3 + 2\text{HCl} \rightarrow \text{CaCl}_2 + \text{CO}_2 + \text{H}_2\text{O} \]Therefore, the moles of CaCO₃ reacted is half that of HCl:\[ \text{moles of CaCO}_3 = \frac{0.0193694 \text{ mol}}{2} = 0.0096847 \text{ mol} \]

Step 6: Calculate Mass of CaCO₃ in Rock

The molar mass of CaCO₃ is \(100.09 \text{ g/mol} \). Calculate the mass of CaCO₃ in the rock:\[ \text{mass of CaCO}_3 = \text{moles of CaCO}_3 \times \text{molar mass of CaCO}_3 = 0.0096847 \, \text{mol} \times 100.09 \, \text{g/mol} \]This gives:\[ \text{mass of CaCO}_3 = 0.9681385 \text{ g} \]

Step 7: Calculate Percentage by Mass of CaCO₃

Finally, calculate the percentage by mass of CaCO₃ in the original rock sample:\[ \text{Percentage of CaCO}_3 = \left(\frac{0.9681385 \, \text{g}}{1.248 \, \text{g}}\right) \times 100\]Calculating this gives:\[ \text{Percentage of CaCO}_3 \approx 77.56\% \]

Key Concepts

Stoichiometry

Stoichiometry is a key concept in chemistry that involves the calculation of reactants and products in chemical reactions. To solve stoichiometry problems, it's crucial to understand the mole ratio of reactants and products, as seen in balanced chemical equations. This helps in determining how much of each substance is needed or produced in a reaction.

For example, in the reaction of calcium carbonate (CaCO₃) with hydrochloric acid (HCl):

• The balanced equation is: \( \text{CaCO}_3 + 2\text{HCl} \rightarrow \text{CaCl}_2 + \text{CO}_2 + \text{H}_2\text{O} \).
• This equation tells us that one mole of CaCO₃ reacts with two moles of HCl.
• By applying stoichiometry, we can calculate the amount of calcium carbonate needed to react completely with a given quantity of HCl.

Understanding stoichiometry allows us to translate the chemical equation into real-world chemical reaction calculations, ensuring that we know exactly how much of each material is involved.

Molarity

Molarity is a measure of the concentration of a solute in a solution. This is typically expressed in moles per liter (mol/L). Molarity is essential for calculating how many moles of a substance are present in a given volume of solution.

To calculate molarity:

• Use the formula \( \text{molarity} = \frac{\text{moles of solute}}{\text{liters of solution}} \).

In the original problem with HCl and NaOH:

• The molarity of HCl is used to find out how many moles of HCl are in 30.00 mL of 1.035 M HCl solution.
• Similarly, molarity helps us determine the moles of NaOH needed to neutralize any excess HCl.

Molarity is a fundamental concept that aids in understanding and performing chemical reaction calculations accurately.

Acid-Base Titration

An acid-base titration is a process used to determine the concentration of an unknown acid or base solution by neutralizing it with a solution of known concentration. This involves a reaction between an acid and a base, typically resulting in the formation of water and a salt.

Here's how it was applied in the exercise:

• In this experiment, HCl (acid) and NaOH (base) are involved in a titration process.
• The initial moles of HCl present were determined, and then any excess HCl not reacting with CaCO₃ was neutralized using NaOH.
• The equivalence point of the titration was used to determine the remaining moles of HCl.

This process helps in accurately determining the amount of reactants involved in chemical reaction calculations.

Percentage Composition

Percentage composition is a mathematical representation of the mass of a component compared to the total mass of the compound. It reflects the proportion of each element or a specific substance within a compound or a mixture.

Here's how it applies to our problem:

• The problem required determining the percentage of calcium carbonate in a limestone rock.
• First, the mass of CaCO₃ was calculated. Then, it was compared with the initial mass of the limestone rock sample to find its percentage composition.
• This is calculated using the formula: \( \text{Percentage by mass} = \left( \frac{\text{mass of part}}{\text{total mass}} \right) \times 100 \).

Understanding percentage composition is crucial for situations where you need to determine the fractional presence of a specific component in a mixture or compound.