Problem 99 - Chemistry: An Atoms-Focused Approach

Question

Table salt contains \(1.54 \mathrm{g}\) of chlorine (as chloride ions) for every \(1.00 \mathrm{g}\) of sodium ions. Which of the following mixtures would react to produce NaCl with no sodium or chlorine left over? a. \(11.0 \mathrm{g}\) of sodium and \(17.0 \mathrm{g}\) of chlorine b. 6.5 g of sodium and \(10.0 \mathrm{g}\) of chlorine c. \(6.5 \mathrm{g}\) of sodium and \(12.0 \mathrm{g}\) of chlorine d. \(6.5 \mathrm{g}\) of sodium and \(8.0 \mathrm{g}\) of chlorine

Step-by-Step Solution

Answer

a) 11.0 g of sodium and 17.0 g of chlorine b) 6.5 g of sodium and 10.0 g of chlorine c) 6.5 g of sodium and 12.0 g of chlorine d) 6.5 g of sodium and 8.0 g of chlorine Answer: b) 6.5 g of sodium and 10.0 g of chlorine.

Step 1: Write down the ratio

The ratio given is that there is \(1.54 \mathrm{g}\) of chlorine for every \(1.00 \mathrm{g}\) of sodium. Therefore, we can represent this as a proportionality: $$ \frac{\mathrm{Mass\, of\, Chlorine}}{1.54}=\frac{\mathrm{Mass\, of\, Sodium}}{1.00} $$

Step 2: Calculate the required mass of chlorine for each mixture

Using the proportionality, we can calculate the required mass of chlorine for each mass of sodium in the given mixtures: a) For \(11.0 \mathrm{g}\) of sodium, $$ \frac{\mathrm{Mass\, of\, Chlorine_{a}}}{1.54}=\frac{11.0}{1.00} \Rightarrow \mathrm{Mass\, of\, Chlorine_{a}}= 11.0\times1.54 = 16.94 \mathrm{g} $$ b) For \(6.5 \mathrm{g}\) of sodium, $$ \frac{\mathrm{Mass\, of\, Chlorine_{b}}}{1.54}=\frac{6.5}{1.00} \Rightarrow \mathrm{Mass\, of\, Chlorine_{b}}= 6.5\times1.54 = 10.01 \mathrm{g} $$ c) For \(6.5 \mathrm{g}\) of sodium, $$ \frac{\mathrm{Mass\, of\, Chlorine_{c}}}{1.54}=\frac{6.5}{1.00} \Rightarrow \mathrm{Mass\, of\, Chlorine_{c}}= 6.5\times1.54 = 10.01 \mathrm{g} $$ d) For \(6.5 \mathrm{g}\) of sodium, $$ \frac{\mathrm{Mass\, of\, Chlorine_{d}}}{1.54}=\frac{6.5}{1.00} \Rightarrow \mathrm{Mass\, of\, Chlorine_{d}}= 6.5\times1.54 = 10.01 \mathrm{g} $$

Step 3: Compare the calculated mass of chlorine with the given mass and select the correct mixture

Compare the calculated mass of chlorine with the given mass of chlorine in each mixture: a) \(\mathrm{Mass\, of\, Chlorine_{a}}= 16.94 \mathrm{g} \neq 17.0 \mathrm{g}\) b) \(\mathrm{Mass\, of\, Chlorine_{b}}= 10.01 \mathrm{g} \approx 10.0 \mathrm{g}\) c) \(\mathrm{Mass\, of\, Chlorine_{c}}= 10.01 \mathrm{g} \neq 12.0 \mathrm{g}\) d) \(\mathrm{Mass\, of\, Chlorine_{d}}= 10.01 \mathrm{g} \neq 8.0 \mathrm{g}\) The only option where the calculated mass of chlorine matches the given mass is option b, where \(6.5 \mathrm{g}\) of sodium and \(10.0 \mathrm{g}\) of chlorine react completely to produce NaCl. Therefore, the correct answer is b.

Key Concepts

Understanding Sodium Chloride

Sodium chloride, commonly known as table salt, is a compound formed from sodium and chlorine. It’s what we add to our food to enhance flavor, but chemically, it's quite fascinating. Sodium chloride is an ionic compound, meaning it consists of positively charged sodium ions and negatively charged chloride ions.
In the formation of sodium chloride, sodium atoms donate one electron to become sodium ions (Na⁺), while chlorine atoms gain an electron to become chloride ions (Cl⁻). This electron transfer creates a strong electrostatic force that holds the ions together, resulting in the solid lattice structure of table salt. Key properties of sodium chloride include:

• High melting point: Sodium chloride melts at about 801°C.
• Good solubility in water: When dissolved, it dissociates into Na⁺ and Cl⁻ ions.
• Conduction of electricity in solution: The ions enable the solution to conduct electricity.

Understanding these properties is essential when studying its chemical behavior and reactions.

Exploring Mass Ratios in Compounds

Mass ratio in chemistry refers to the ratio of the masses of elements in a compound. It’s an important stoichiometric concept that helps determine the amount of each element present. In the case of sodium chloride, the mass ratio is given as 1.54 grams of chlorine for every 1 gram of sodium.
This ratio helps predict how much of each element is required to form a compound without any excess, as was illustrated in the exercise with various amounts of sodium and chlorine. By comparing the given chlorine mass with the mass calculated using the ratio, we can identify which amounts react perfectly without any leftover.Steps to determine the required mass:

• Use the formula \( \frac{\text{Mass of Chlorine}}{1.54} = \frac{\text{Mass of Sodium}}{1.00} \).
• Calculate the theoretical mass of chlorine needed for the given mass of sodium.
• Compare this with the actual mass available to check for a match.

Understanding mass ratios ensures precise chemical calculations and helps avoid waste of chemicals during reactions.

Essentials of Chemical Reactions

Chemical reactions involve the transformation of reactants into products through the breaking and forming of chemical bonds. In the reaction to produce sodium chloride, sodium (Na) reacts with chlorine (Cl₂) in a straightforward way. The balanced equation for this reaction is:
\[ 2\text{Na} + \text{Cl}_2 \rightarrow 2\text{NaCl} \]This equation represents 2 atoms of sodium reacting with 1 molecule of chlorine gas to form 2 units of sodium chloride.Let's break down what happens in this reaction:

• Each sodium atom gives up one electron, becoming a Na⁺ ion.
• Each chlorine atom accepts one electron, forming Cl⁻ ions.
• The ionic bonds formed between Na⁺ and Cl⁻ produce a stable compound, NaCl.

Understanding the principles of chemical reactions, like balancing equations and electron transfer, is crucial in predicting reaction outcomes and ensuring that reactants convert completely into products. This knowledge is fundamental in stoichiometry and the efficient use of materials in chemical processes.