Question: Name Section Experiment 8 ADVANCE STUDY ASSIGNMENT 1. A 0.5118-G Sample Of Caco, Is Dissolved In 6 M HCl, And The
The correct answer and explanation is:
To answer your question, it seems like you are referring to an experiment involving the dissolution of calcium carbonate (CaCO₃) in hydrochloric acid (HCl). Here’s the correct interpretation and the explanation:
Correct Answer:
The balanced chemical reaction between calcium carbonate and hydrochloric acid is: CaCO₃ (s)+2HCl (aq)→CaCl₂ (aq)+H₂O (l)+CO₂ (g)\text{CaCO₃ (s)} + 2 \text{HCl (aq)} \rightarrow \text{CaCl₂ (aq)} + \text{H₂O (l)} + \text{CO₂ (g)}
In this reaction, calcium carbonate reacts with hydrochloric acid to form calcium chloride, water, and carbon dioxide gas.
Explanation:
In this experiment, a 0.5118 g sample of calcium carbonate (CaCO₃) is dissolved in 6 M hydrochloric acid (HCl). This dissolution reaction is a typical acid-base reaction, where the acid (HCl) reacts with the base (CaCO₃) to produce calcium chloride (CaCl₂), water (H₂O), and carbon dioxide (CO₂) gas.
The first step is to write the balanced chemical equation. When calcium carbonate is added to hydrochloric acid, it reacts as follows: CaCO₃ (s)+2HCl (aq)→CaCl₂ (aq)+H₂O (l)+CO₂ (g)\text{CaCO₃ (s)} + 2 \text{HCl (aq)} \rightarrow \text{CaCl₂ (aq)} + \text{H₂O (l)} + \text{CO₂ (g)}
This equation is balanced because the number of atoms of each element is the same on both sides of the equation. Specifically, one calcium atom (Ca) reacts with two chloride ions (Cl⁻), and the carbonate (CO₃²⁻) reacts with the acid to release carbon dioxide gas and form water.
To understand the stoichiometry of the reaction, we calculate the moles of calcium carbonate in the sample. Using the molar mass of CaCO₃ (which is 100.09 g/mol), we can find the number of moles of CaCO₃ in the 0.5118 g sample: moles of CaCO₃=0.5118 g100.09 g/mol=0.00511 mol\text{moles of CaCO₃} = \frac{0.5118 \text{ g}}{100.09 \text{ g/mol}} = 0.00511 \text{ mol}
From the balanced equation, we know that 1 mole of CaCO₃ reacts with 2 moles of HCl. Therefore, we can calculate the amount of HCl needed for complete reaction and the volume of gas produced. This experiment is useful in understanding acid-base reactions and the stoichiometry involved.