Question

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

Short answer

The percent by mass of calcium carbonate in the limestone rock is approximately $77.61\mathrm{\%}$.

Step-by-step solution

Calculate the moles of HCl initially present

To find the moles of HCl initially present, we'll use the concentration and volume of the solution: moles of HCl = concentration × volume moles of HCl = 1.035 mol/L × 30.00 mL × (1 L / 1000 mL) moles of HCl = 0.03105 mol

Calculate the moles of NaOH used for neutralization

To find the moles of NaOH used for neutralization, we'll use the concentration and volume of the NaOH solution: moles of NaOH = concentration × volume moles of NaOH = 1.010 mol/L × 11.56 mL × (1 L / 1000 mL) moles of NaOH = 0.01168 mol

Calculate the moles of HCl that reacted with CaCO3

Since the NaOH reacts with the excess HCl in 1:1 ratio, the moles of HCl that reacted with CaCO3 are: moles of HCl reacted with CaCO3 = moles of HCl initially present – moles of NaOH moles of HCl reacted with CaCO3 = 0.03105 mol – 0.01168 mol moles of HCl reacted with CaCO3 = 0.01937 mol

Calculate the moles of CaCO3

From the balanced chemical equation, we can see that 1 mole of CaCO3 reacts with 2 moles of HCl: CaCO3 + 2HCl → CaCl2 + H2O + CO2 Thus, we can find the moles of CaCO3 by dividing the moles of HCl that reacted with CaCO3 by 2: moles of CaCO3 = moles of HCl reacted with CaCO3 / 2 moles of CaCO3 = 0.01937 mol / 2 moles of CaCO3 = 0.009685 mol

Calculate the mass of CaCO3

To find the mass of CaCO3, we'll use its molar mass (100.09 g/mol) and the moles of CaCO3: mass of CaCO3 = moles of CaCO3 × molar mass of CaCO3 mass of CaCO3 = 0.009685 mol × 100.09 g/mol mass of CaCO3 = 0.9689 g

Calculate the percent by mass of CaCO3 in the rock

Now, we will find the percent by mass of CaCO3 in the rock using the mass of CaCO3 and the mass of the rock: percent by mass of CaCO3 = (mass of CaCO3 / mass of the rock) × 100 percent by mass of CaCO3 = (0.9689 g / 1.248 g) × 100 percent by mass of CaCO3 = 77.61 % Therefore, the percent by mass of calcium carbonate in the rock is approximately 77.61%.

Key concepts

Stoichiometry

Stoichiometry is a fundamental concept in chemistry that deals with the quantitative relationships between reactants and products in a chemical reaction. It allows chemists to predict the amounts of substances consumed and produced in any given reaction.

For instance, in the step-by-step solution of our exercise, stoichiometry was used to determine the number of moles of calcium carbonate (CaCO3) that could be formed from the reaction between HCl and CaCO3. Knowing that HCl reacts with CaCO3 in a 2:1 ratio, stoichiometry provides the framework to deduce the quantity of each reactant needed to produce a desired amount of product.

In educational exercises, errors often arise when students fail to balance chemical equations or misunderstand the molar ratios of reactants and products. A balanced chemical equation is crucial for correct stoichiometric calculations, as it represents the foundation upon which these calculations are made.

Molarity

Molarity is a measure of the concentration of a solute in a solution, or in other words, how much of a substance is contained within a certain volume of liquid. It is defined as the number of moles of solute per liter of solution. The formula for calculating molarity is:
$$\text{Molarity (M)}=\frac{\text{moles of solute}}{\text{liters of solution}}$$
In our solution, molarity plays a critical role in determining the starting quantities of HCl and NaOH. By using the provided concentrations and volumes of the solutions, students can calculate the moles of HCl and NaOH, which is the first step in understanding the overall stoichiometry of the neutralization reaction.

Common mistakes often include confusing molarity with molality or incorrectly converting milliliters to liters. To help prevent these errors, students should be encouraged to consistently use units and conversion factors in their calculations.

Neutralization Reaction

A neutralization reaction is a type of chemical reaction in which an acid and a base react to form water and a salt. This process typically involves the transfer of a proton (H+) from the acid to the hydroxide ion (OH-) from the base, resulting in the formation of water (H2O).

In the given exercise, a neutralization reaction occurs when the excess HCl is treated with NaOH. This specific reaction is characterized by a 1:1 molar ratio, meaning that one mole of acid reacts with one mole of base. Understanding the underlying stoichiometry allows the calculation of the moles of acid that reacted with the limestone.

• Identify the reactants and products in the reaction
• Use the 1:1 ratio to relate the moles of NaOH used to the moles of excess HCl neutralized
• Infer the moles of HCl that participated in the original reaction with limestone (CaCO3)

Correctly interpreting a neutralization reaction is key to solving problems related to percent composition by mass, as seen in our exercise. It's essential for students to recognize the indicators of a reaction reaching neutralization, which can be observed through pH changes or the amounts of reactants used.