Problem 113 - Chemistry and Chemical Reactivity

Question

A You place \(2.56 \mathrm{g}\) of \(\mathrm{CaCO}_{3}\) in a beaker containing \(250 .\) mL of \(0.125 \mathrm{M} \mathrm{HCl}\). When the reaction has ceased, does any calcium carbonate remain? What mass of \(\mathrm{CaCl}_{2}\) can be produced? \(\mathrm{CaCO}_{3}(\mathrm{s})+2 \mathrm{HCl}(\mathrm{aq}) \rightarrow\) $$ \mathrm{CaCl}_{2}(\mathrm{aq})+\mathrm{CO}_{2}(\mathrm{g})+\mathrm{H}_{2} \mathrm{O}(\ell) $$

Step-by-Step Solution

Answer

1.00 g of \( \mathrm{CaCO}_3 \) remains; 1.73 g of \( \mathrm{CaCl}_2 \) is produced.

Step 1: Determine Moles of Reactants

Calculate the moles of \( \mathrm{CaCO}_3 \) using its molar mass. The molar mass of \( \mathrm{CaCO}_3 \) is approximately \( 100.09 \, \mathrm{g/mol} \). Thus, \[ \text{Moles of } \mathrm{CaCO}_3 = \frac{2.56 \, \mathrm{g}}{100.09 \, \mathrm{g/mol}} \approx 0.0256 \, \mathrm{mol} \]Next, calculate the moles of \( \mathrm{HCl} \) in the solution. The concentration of \( \mathrm{HCl} \) is \( 0.125 \, \mathrm{M} \) and the volume is \( 0.250 \, \mathrm{L} \).\[ \text{Moles of } \mathrm{HCl} = 0.125 \, \mathrm{mol/L} \times 0.250 \, \mathrm{L} = 0.03125 \, \mathrm{mol} \]

Step 2: Identify the Limiting Reactant

The balanced chemical equation shows that 1 mole of \( \mathrm{CaCO}_3 \) reacts with 2 moles of \( \mathrm{HCl} \). We have 0.0256 moles of \( \mathrm{CaCO}_3 \) and need 0.0512 moles of \( \mathrm{HCl} \) (since 0.0256 mol \( \times 2 \) = 0.0512 mol). However, there are only 0.03125 moles of \( \mathrm{HCl} \) available, which is less than required.Thus, \( \mathrm{HCl} \) is the limiting reactant.

Step 3: Calculate Remaining \( \mathrm{CaCO}_3 \)

Since \( \mathrm{HCl} \) is the limiting reactant, not all of the \( \mathrm{CaCO}_3 \) will react. Calculate the moles of \( \mathrm{CaCO}_3 \) that react using the total moles of \( \mathrm{HCl} \).\[ \text{Moles of } \mathrm{CaCO}_3 \text{ that react} = \frac{0.03125 \, \mathrm{mol} \text{ of } \mathrm{HCl}}{2} = 0.015625 \, \mathrm{mol} \]Subtract from the initial moles to determine remaining \( \mathrm{CaCO}_3 \).\[ 0.0256 \, \mathrm{mol} - 0.015625 \, \mathrm{mol} = 0.009975 \, \mathrm{mol} \]

Step 4: Calculate Mass of Remaining \( \mathrm{CaCO}_3 \)

Convert the remaining moles of \( \mathrm{CaCO}_3 \) back to grams using its molar mass.\[ \text{Mass of remaining } \mathrm{CaCO}_3 = 0.009975 \, \mathrm{mol} \times 100.09 \, \mathrm{g/mol} \approx 1.00 \, \mathrm{g} \]

Step 5: Calculate Mass of \( \mathrm{CaCl}_2 \) Produced

Calculate the moles of \( \mathrm{CaCl}_2 \) produced, which is equal to the moles of \( \mathrm{CaCO}_3 \) that reacted (since their ratio is 1:1).\[ 0.015625 \, \mathrm{mol} \text{ of } \mathrm{CaCl}_2 \]Convert this to grams using the molar mass of \( \mathrm{CaCl}_2 \) (~110.98 g/mol).\[ \text{Mass of } \mathrm{CaCl}_2 = 0.015625 \, \mathrm{mol} \times 110.98 \, \mathrm{g/mol} \approx 1.73 \, \mathrm{g} \]

Key Concepts

Limiting Reactant

In a chemical reaction, the limiting reactant is the substance that gets completely used up first, thus determining the amount of product formed. It limits the extent of the reaction. Identifying the limiting reactant is crucial when multiple reactants are involved. This ensures we don't overestimate the amount of product that can be formed.

To find the limiting reactant, it's important to consult the balanced chemical equation. This equation provides the stoichiometric ratios necessary to compare the reactants. For example, in the equation provided, 1 mole of \( \mathrm{CaCO}_3 \) would need 2 moles of \( \mathrm{HCl} \). By calculating the number of moles available from each reactant and comparing with the stoichiometric ratios, you can determine the limiting reactant.

• Identify the moles required for each reactant based on the equation.
• Compare available moles with required moles.
• The reactant with fewer moles than required becomes the limiting reactant.

This concept helps us not only predict the amount of product but also calculate the unused reactants left behind.

Molar Mass

The molar mass of a substance is the mass of one mole of its particles. It is usually expressed in grams per mole \( \mathrm{g/mol} \). Molar mass serves as a bridge between mass and moles, enabling us to convert between the two.

To calculate molar mass, sum up the atomic masses of all atoms in a chemical formula based on the periodic table. For instance, the molar mass of \( \mathrm{CaCO}_3 \) is calculated as follows:

• Calcium (Ca): 40.08 g/mol
• Carbon (C): 12.01 g/mol
• Oxygen (O): \(3 \times 16.00 \) g/mol = 48.00 g/mol

Adding these results in a molar mass of 100.09 g/mol for \( \mathrm{CaCO}_3 \).
Understanding molar mass is essential for performing moles to grams and grams to moles calculations in any stoichiometric problem.

Chemical Reactions

Chemical reactions involve the transformation of reactants into products. Reactants are substances that start a reaction, while products are substances formed as a result. A chemical equation represents the rearrangement of atoms to form new substances.

In our example, the reaction is:
\[\mathrm{CaCO}_3(\mathrm{s})+2 \mathrm{HCl}(\mathrm{aq}) \rightarrow\ \mathrm{CaCl}_2(\mathrm{aq})+\mathrm{CO}_2(\mathrm{g})+\mathrm{H}_2\mathrm{O}(\ell)\]
This reaction states that calcium carbonate and hydrochloric acid react to produce calcium chloride, carbon dioxide, and water. The coefficients help identify stoichiometric ratios, which are crucial when calculating how much of each reactant is needed or product produced.

• Ensure the equation is balanced.
• Identify reactants and products clearly.
• Understand that coefficients indicate the ratio in which substances react and form.

Such understanding aids in deeper interpretations of reactions, whether in theoretical problem-solving or real-world applications.

Moles Calculation

A mole is a fundamental unit in chemistry representing \( 6.022 \times 10^{23} \) particles (atoms, molecules, ions, etc.). Moles simplify the process of converting masses into quantities of substance, using Avogadro's number.

The formula to calculate moles from mass is:
\[ \text{Moles} = \frac{\text{mass in grams}}{\text{molar mass in g/mol}} \]
In stoichiometry, moles are used to relate masses of reactants involved to masses of products formed. For instance, to find the moles of \( \mathrm{CaCO}_3 \) in 2.56 g, we calculate:
\[\text{Moles of } \mathrm{CaCO}_3 = \frac{2.56 \, \mathrm{g}}{100.09 \, \mathrm{g/mol}} \approx 0.0256 \, \mathrm{mol}\]
This calculation allows us to determine how much \( \mathrm{CaCO}_3 \) participates in forming products, aiding in further analyses such as finding limiting reactants or excess amounts. Accurate mole calculations are foundational in the stoichiometric evaluation of chemical reactions.