Problem 113 - Chemistry and Chemical Reactivity

Question

You place \(2.56 \mathrm{g}\) of \(\mathrm{CaCO}_{3}\) in a beaker containing \(250 .\) mL of \(0.125 \mathrm{M}\) 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

Yes, 0.9975 g of CaCO3 remains. 1.733 g of CaCl2 can be produced.

Step 1: Calculate Moles of Reactants

First, we need to calculate the number of moles of \( \mathrm{CaCO}_{3} \) and \( \mathrm{HCl} \). The molar mass of \( \mathrm{CaCO}_{3} \) is approximately \( 100.09 \text{ g/mol} \). The moles of \( \mathrm{CaCO}_{3} \) are calculated as follows:\[\text{Moles of } \mathrm{CaCO}_{3} = \frac{2.56 \text{ g}}{100.09 \text{ g/mol}} = 0.0256 \text{ moles}\]For \( \mathrm{HCl} \), using the concentration and volume, we find its moles:\[\text{Moles of } \mathrm{HCl} = 0.125 \text{ M} \times 0.250 \text{ L} = 0.03125 \text{ moles}\]

Step 2: Determine Limiting Reactant

Using the stoichiometry of the balanced equation \( \mathrm{CaCO}_{3} + 2 \mathrm{HCl} \rightarrow \mathrm{CaCl}_{2} + \mathrm{CO}_{2} + \mathrm{H}_{2}\mathrm{O} \), we see that 1 mole of \( \mathrm{CaCO}_{3} \) reacts with 2 moles of \( \mathrm{HCl} \). Thus, the complete reaction of 0.0256 mol of \( \mathrm{CaCO}_{3} \) would require:\[2 \times 0.0256 = 0.0512 \text{ moles of } \mathrm{HCl}\]Since we only have 0.03125 moles of \( \mathrm{HCl} \), \( \mathrm{HCl} \) is the limiting reactant, and \( \mathrm{CaCO}_{3} \) is in excess.

Step 3: Determine Remaining \( \mathrm{CaCO}_{3} \)

Since \( \mathrm{HCl} \) is limiting, it will be consumed completely. Find the moles of \( \mathrm{CaCO}_{3} \) that reacted:\[\text{Moles of } \mathrm{CaCO}_{3} \text{ reacted} = \frac{0.03125}{2} = 0.015625 \text{ moles}\]The moles of \( \mathrm{CaCO}_{3} \) remaining are:\[0.0256 - 0.015625 = 0.009975 \text{ moles}\]Thus, some \( \mathrm{CaCO}_{3} \) remains.

Step 4: Calculate Mass of \( \mathrm{CaCl}_{2} \) Produced

Calculate the moles of \( \mathrm{CaCl}_{2} \) produced, which is equal to the moles of \( \mathrm{CaCO}_{3} \) reacted, since they have a 1:1 ratio:\[0.015625 \text{ moles of } \mathrm{CaCl}_{2}\]Calculate the mass of \( \mathrm{CaCl}_{2} \) produced using its molar mass (110.98 g/mol):\[\text{Mass of } \mathrm{CaCl}_{2} = 0.015625 \times 110.98 = 1.733 \text{ g}\]

Key Concepts

Limiting Reactant

When it comes to chemical reactions, identifying the limiting reactant is crucial. This is the reactant that runs out first, preventing the reaction from proceeding any further. In our exercise, the balanced chemical equation shows that one mole of \( \text{CaCO}_3 \) reacts with two moles of \( \text{HCl} \). Given 0.0256 moles of \( \text{CaCO}_3 \) and 0.03125 moles of \( \text{HCl} \), we need to determine which one limits the reaction.

• 0.0256 moles of \( \text{CaCO}_3 \) requires 0.0512 moles of \( \text{HCl} \). However, only 0.03125 moles of \( \text{HCl} \) are available.
• This means \( \text{HCl} \) is the limiting reactant, as there isn't enough to react with all the \( \text{CaCO}_3 \) present.

To find out if any \( \text{CaCO}_3 \) is left, it's important to calculate the amount that reacts. Since \( \text{HCl} \) is limiting, 0.015625 moles of \( \text{CaCO}_3 \) react, and 0.009975 moles are leftover. The presence of excess \( \text{CaCO}_3 \) confirms its non-limiting role in the reaction.

Chemical Reactions

Chemical reactions are processes where substances change to form new products. In our scenario, calcium carbonate (\( \text{CaCO}_3 \)) reacts with hydrochloric acid (\( \text{HCl} \)) to produce calcium chloride (\( \text{CaCl}_2 \)), carbon dioxide (\( \text{CO}_2 \)), and water (\( \text{H}_2\text{O} \)). Understanding the roles of each component is essential:

• \( \text{CaCO}_3 \) acts as a base, reacting with the acid \( \text{HCl} \).
• The products \( \text{CaCl}_2 \), \( \text{CO}_2 \), and \( \text{H}_2\text{O} \) signify a typical acid-carbonate reaction.

The balanced equation helps to ensure the conservation of mass, where the number of each type of atom on the reactant side equals the number on the product side. This stoichiometry involves paying attention to coefficients in the equation, dictating how many moles of each reactant are needed for the reaction. Understanding these principles is key to predicting the outcomes and leftovers in such reactions.

Molar Mass Calculation

Knowing how to calculate molar mass is a foundational step in stoichiometry. Molar mass is the mass of one mole of a substance and is expressed in grams per mole. It's crucial for converting between mass and moles, which is essential in determining limiting reactants and predicting product yields.
In this exercise, we have:

• For \( \text{CaCO}_3 \), the molar mass is approximately 100.09 g/mol. It includes the atomic masses of calcium (Ca), carbon (C), and three oxygens (O).
• For \( \text{CaCl}_2 \), the molar mass is approximately 110.98 g/mol.

By dividing the given mass by the molar mass, you can calculate the number of moles. For example, with 2.56 g of \( \text{CaCO}_3 \), by using the formula \( \frac{\text{mass}}{\text{molar mass}} \), we find it equals 0.0256 moles. This accuracy in calculating moles is critical in determining how much of a reactant will react or remain, as well as the amount of product formed.