Problem 110 - General Chemistry: Principles and Modern Applications

Question

Solid calcium carbonate, \(\mathrm{CaCO}_{3}(\mathrm{s}),\) reacts with \(\mathrm{HCl}(\mathrm{aq})\) to form \(\mathrm{H}_{2} \mathrm{O}, \mathrm{CaCl}_{2}(\mathrm{aq}),\) and \(\mathrm{CO}_{2}(\mathrm{g}) .\) If a \(45.0 \mathrm{g}\) sample of \(\mathrm{CaCO}_{3}(\mathrm{s})\) is added to \(1.25 \mathrm{L}\) of \(\mathrm{HCl}(\mathrm{aq})\) that is \(25.7 \% \mathrm{HCl}\) by mass \((d=1.13 \mathrm{g} / \mathrm{mL})\) what will be the molarity of \(\mathrm{HCl}\) in the solution after the reaction is completed? Assume that the solution volume remains constant.

Step-by-Step Solution

Answer

After the reaction is completed, the molarity of the HCl in the solution is 7.032 M.

Step 1: Calculation of Initial Moles of HCl

First, find the total mass of the solution, which is \(\text{Volume} \times \(\text{Density} = 1.25 \, \text{L} \times 1.13 \, \text{g/mL} = 1375 \, \text{g}\). Then, determine the mass of HCl in the solution using the mass percent given, which is \(0.257 \times 1375 \, \text{g} = 353.625 \, \text{g HCl}\). The molar mass of HCl is approximately 36.5 g/mol. Thus, initial moles of HCl can be calculated as \( \frac{353.625 \, \text{g}}{36.5 \, \text{g/mol}} = 9.69 \, \text{moles}\).

Step 2: Calculation of Moles of CaCO3

Next, figure out the moles of CaCO3, using its given mass and molar mass, which is approximately 100 g/mol. So, the number moles of CaCO3 = \( \frac{45 \, \text{g}}{100 \, \text{g/mol}} = 0.45 \, \text{moles}\).

Step 3: Determination of Limiting Reactant

From the balanced chemical equation \(\text{CaCO}_3 + 2\text{HCl} \rightarrow \text{H}_2\text{O} + \text{CaCl}_2 + \text{CO}_2\), it can be seen that one mole of \(\text{CaCO}_3\) reacts with two moles of HCl. Therefore, in order to react with 0.45 moles of \(\text{CaCO}_3\), we would need \(0.45 \times 2 = 0.9\) moles of HCl. But 9.69 moles are available, so HCl is in excess, and \(\text{CaCO}_3\) is limiting. Determination of the limiting reactant is crucial as it limits the progression of the reaction.

Step 4: Calculation of Final moles of HCl

As \(\text{CaCO}_3\) is the limiting reactant, all of it will react with HCl. Knowing the stoichiometry, 0.9 moles of HCl will react with \(\text{CaCO}_3\). Thus, subtract the moles of HCl reacted from the initial moles. Therefore, final moles of HCl = Initial moles - reacted moles = 9.69 - 0.9 = 8.79 moles.

Step 5: Calculation of Final Molarity of HCl

Since the volume of the solution remains constant (1.25 L), the final molarity of HCl = moles of HCl / volume of the solution in liters. Therefore, Molarity = \(\frac{8.79 \, \text{moles}}{1.25 \, \text{L}} = 7.032 \, \text{M}\).

Key Concepts

Molarity Calculation

When we talk about molarity, we are referring to a way of expressing the concentration of a solution. Molarity is defined as the number of moles of a solute (in this case, hydrochloric acid, HCl) per liter of solution. This gives us insight into how much substance is present in a volume, which is crucial in reactions.

To calculate molarity, we follow a straightforward formula:

• Molarity (M) = moles of solute / volume of solution in liters.

In the given exercise, we first calculated the initial moles of HCl by using the mass and density of the solution. From there, by knowing the initial moles and the total volume of the solution, we were able to determine the molarity of HCl after the reaction. In practice, this helps us to understand how concentrated the acid is throughout and after the chemical process.

Balanced Chemical Equation

Chemical reactions must abide by the law of conservation of mass, which tells us that mass can neither be created nor destroyed in a chemical reaction. This leads to the necessity of balanced chemical equations. A balanced chemical equation has the same number of atoms of each element on both sides of the equation.

In the given reaction, the balanced equation was:

• \(\mathrm{CaCO}_3 (\mathrm{s}) + 2 \mathrm{HCl} (\mathrm{aq}) \rightarrow \mathrm{H}_2\mathrm{O} + \mathrm{CaCl}_2 (\mathrm{aq}) + \mathrm{CO}_2 (\mathrm{g})\)

For every molecule of calcium carbonate \(\mathrm{CaCO}_3\), two molecules of hydrochloric acid \(\mathrm{HCl}\) are required. By balancing the equation, we ensured that the reaction could proceed correctly and that we could accurately predict the amount of products formed based on the stoichiometry of the reactants and products.

Stoichiometry

Stoichiometry is the quantitative study of reactants and products in a chemical reaction. It involves using a balanced chemical equation to find the proportion of reactants needed and the amount of products that will be generated.

In the exercise, stoichiometry was essential to figure out the limiting reactant. By knowing the molar amounts of \(\mathrm{CaCO}_3\) and \(\mathrm{HCl}\), and using the stoichiometric coefficients from the balanced equation, we determined that \(\mathrm{CaCO}_3\) was the limiting reactant since it would be completely consumed.Here's a quick breakdown of why this is important:

• It helps predict the extent of the reaction based on initial quantities.
• Allows estimation of remaining reactants after the reaction is complete.
• Helps calculate the yield of the product formed.

Thus, stoichiometry enables chemists to make precise calculations about how a reaction progresses, ensuring accuracy in experiments and industrial processes.