NaCl Formation Calculation: How Much NaCl From 0.25g Na And 0.39g Cl₂
To determine the mass of NaCl formed when 0.25 g of sodium (Na) reacts completely with 0.39 g of chlorine (Cl₂), we need to delve into the stoichiometry of the reaction. Stoichiometry is the branch of chemistry that involves quantitative relationships between reactants and products in chemical reactions. This understanding is crucial for various applications, from industrial chemical production to laboratory experiments. In this article, we will explore the step-by-step process of calculating the mass of NaCl formed, shedding light on the underlying principles and ensuring a clear understanding of the solution. First, we will write the balanced chemical equation for the reaction. This balanced equation will serve as the foundation for our calculations, providing the necessary molar ratios between reactants and products. Then, we will convert the given masses of reactants into moles using their respective molar masses. This conversion is essential as chemical reactions occur based on molar ratios, not mass ratios. Next, we will identify the limiting reactant, which is the reactant that is completely consumed in the reaction and determines the maximum amount of product that can be formed. This is a critical step because the amount of NaCl formed will be dictated by the limiting reactant. Once we have identified the limiting reactant, we will use the stoichiometric ratio from the balanced equation to calculate the moles of NaCl produced. This calculation involves using the mole ratio between the limiting reactant and NaCl. Finally, we will convert the moles of NaCl formed into grams using the molar mass of NaCl. This final step will provide us with the mass of NaCl formed in the reaction. By meticulously following these steps, we will arrive at the correct answer and gain a deeper understanding of stoichiometric calculations in chemistry.
Balanced Chemical Equation
The reaction between sodium (Na) and chlorine (Cl₂) to form sodium chloride (NaCl) is a classic example of a synthesis reaction. It is fundamental to understanding chemical reactions and stoichiometry. The reaction involves the combination of two elements to form a compound, and it's essential to have the balanced chemical equation to accurately determine the quantitative relationships between the reactants and products. The unbalanced equation is:
Na + Cl₂ → NaCl
To balance this equation, we need to ensure that the number of atoms of each element is the same on both sides of the equation. We have one sodium atom on the left and one on the right, which is balanced. However, we have two chlorine atoms on the left and only one on the right. To balance the chlorine atoms, we place a coefficient of 2 in front of NaCl:
Na + Cl₂ → 2 NaCl
Now, we have two chlorine atoms on both sides, but we have two sodium atoms on the right and only one on the left. To balance the sodium atoms, we place a coefficient of 2 in front of Na:
2 Na + Cl₂ → 2 NaCl
This is the balanced chemical equation. It tells us that two moles of sodium react with one mole of chlorine gas to produce two moles of sodium chloride. This balanced equation is crucial for our stoichiometric calculations. It provides the mole ratios needed to determine the amount of product formed from the given amounts of reactants. The coefficients in the balanced equation represent the molar ratios, which are essential for converting between the amounts of reactants and products. In this case, the molar ratio between Na and NaCl is 2:2, which simplifies to 1:1. The molar ratio between Cl₂ and NaCl is 1:2. These ratios will be used later to determine the limiting reactant and the amount of NaCl formed. Understanding and correctly balancing chemical equations is a fundamental skill in chemistry, as it forms the basis for stoichiometric calculations and predicting the outcomes of chemical reactions.
Converting Grams to Moles
In stoichiometric calculations, converting grams to moles is a crucial step. This conversion allows us to work with the molar ratios provided by the balanced chemical equation. Moles are the standard unit for measuring the amount of a substance in chemistry, and reactions occur based on molar ratios, not mass ratios. To convert grams to moles, we use the molar mass of the substance. The molar mass is the mass of one mole of a substance and is expressed in grams per mole (g/mol). The molar mass can be calculated by summing the atomic masses of all the atoms in the chemical formula, which are typically found on the periodic table. For sodium (Na), the atomic mass is approximately 22.99 g/mol. This means that one mole of sodium has a mass of 22.99 grams. To convert 0.25 g of Na to moles, we use the formula:
Moles = Mass / Molar Mass
Moles of Na = 0.25 g / 22.99 g/mol ≈ 0.0109 moles
For chlorine (Cl₂), the atomic mass of a single chlorine atom is approximately 35.45 g/mol. Since chlorine exists as a diatomic molecule (Cl₂), the molar mass of Cl₂ is twice the atomic mass of chlorine:
Molar mass of Cl₂ = 2 × 35.45 g/mol = 70.90 g/mol
To convert 0.39 g of Cl₂ to moles, we use the same formula:
Moles of Cl₂ = 0.39 g / 70.90 g/mol ≈ 0.0055 moles
Now we have the amounts of both reactants in moles. These values will be used to determine the limiting reactant, which is the reactant that is completely consumed in the reaction and determines the maximum amount of product that can be formed. Converting grams to moles is a fundamental skill in stoichiometry, and it is essential for accurately calculating the amounts of reactants and products involved in chemical reactions. Understanding this conversion allows us to move from the macroscopic scale (grams) to the microscopic scale (moles), which is necessary for understanding chemical reactions at a molecular level. In the next step, we will use these mole values to identify the limiting reactant and then calculate the amount of NaCl formed.
Identifying the Limiting Reactant
Identifying the limiting reactant is a critical step in stoichiometry. The limiting reactant is the reactant that is completely consumed in a chemical reaction. It determines the maximum amount of product that can be formed. The other reactant(s) are considered to be in excess because there will be some amount of them left over after the reaction is complete. To identify the limiting reactant, we compare the mole ratios of the reactants to the stoichiometric ratios from the balanced chemical equation. The balanced chemical equation for the reaction between sodium and chlorine is:
2 Na + Cl₂ → 2 NaCl
From the balanced equation, we see that 2 moles of Na react with 1 mole of Cl₂. We have already calculated the moles of each reactant:
Moles of Na = 0.0109 moles Moles of Cl₂ = 0.0055 moles
To determine the limiting reactant, we can compare the mole ratio of the reactants to the stoichiometric ratio. We can divide the moles of each reactant by its respective coefficient in the balanced equation:
For Na: 0.0109 moles / 2 = 0.00545 For Cl₂: 0.0055 moles / 1 = 0.0055
The reactant with the smallest value is the limiting reactant. In this case, Na has a smaller value (0.00545) compared to Cl₂ (0.0055). Therefore, Na is the limiting reactant. This means that all 0.0109 moles of Na will be consumed in the reaction, and the amount of NaCl formed will be determined by the amount of Na available. Cl₂ is in excess, so there will be some Cl₂ left over after the reaction is complete. Another way to identify the limiting reactant is to calculate the amount of product that can be formed from each reactant. We can use the stoichiometric ratios to determine how many moles of NaCl can be formed from the given moles of Na and Cl₂. From the balanced equation, 2 moles of Na produce 2 moles of NaCl, and 1 mole of Cl₂ produces 2 moles of NaCl. Using Na:
Moles of NaCl = (0.0109 moles Na) × (2 moles NaCl / 2 moles Na) = 0.0109 moles NaCl
Using Cl₂:
Moles of NaCl = (0.0055 moles Cl₂) × (2 moles NaCl / 1 mole Cl₂) = 0.0110 moles NaCl
The reactant that produces the smaller amount of product is the limiting reactant. In this case, Na produces 0.0109 moles of NaCl, while Cl₂ produces 0.0110 moles of NaCl. Therefore, Na is the limiting reactant. Identifying the limiting reactant is crucial because it allows us to accurately calculate the amount of product formed. The amount of product formed cannot exceed the amount that can be produced from the limiting reactant. In the next step, we will use the moles of the limiting reactant to calculate the moles of NaCl formed and then convert it to grams.
Calculating Moles of NaCl Formed
Now that we have identified sodium (Na) as the limiting reactant, we can calculate the moles of sodium chloride (NaCl) formed in the reaction. This calculation is based on the stoichiometric ratio between the limiting reactant and the product, as given by the balanced chemical equation:
2 Na + Cl₂ → 2 NaCl
The balanced equation tells us that 2 moles of Na react to produce 2 moles of NaCl. This means the mole ratio between Na and NaCl is 2:2, which simplifies to 1:1. To calculate the moles of NaCl formed, we use the moles of the limiting reactant (Na) and the mole ratio:
Moles of NaCl = Moles of Na × (Moles of NaCl / Moles of Na)
We have 0.0109 moles of Na (the limiting reactant). Plugging this value into the equation:
Moles of NaCl = 0.0109 moles Na × (2 moles NaCl / 2 moles Na) Moles of NaCl = 0.0109 moles Na × (1 mole NaCl / 1 mole Na) Moles of NaCl = 0.0109 moles
Therefore, 0.0109 moles of NaCl are formed in the reaction. This calculation is straightforward because the mole ratio between Na and NaCl is 1:1. If the mole ratio were different, we would simply multiply the moles of the limiting reactant by the appropriate mole ratio to find the moles of product formed. For example, if the balanced equation showed that 1 mole of the limiting reactant produced 2 moles of the product, we would multiply the moles of the limiting reactant by 2. Calculating the moles of product formed is a critical step in stoichiometry because it allows us to determine the theoretical yield of the reaction. The theoretical yield is the maximum amount of product that can be formed from the given amounts of reactants, assuming the reaction goes to completion and there are no losses. In reality, the actual yield of a reaction may be less than the theoretical yield due to factors such as incomplete reactions, side reactions, or losses during product isolation and purification. However, the theoretical yield provides a benchmark for assessing the efficiency of a reaction. In the next step, we will convert the moles of NaCl formed into grams to determine the mass of NaCl produced in the reaction.
Converting Moles of NaCl to Grams
To determine the mass of NaCl formed, we need to convert the moles of NaCl we calculated in the previous step (0.0109 moles) into grams. This conversion is done using the molar mass of NaCl. The molar mass is the mass of one mole of a substance, expressed in grams per mole (g/mol). To calculate the molar mass of NaCl, we add the atomic masses of sodium (Na) and chlorine (Cl) from the periodic table:
Atomic mass of Na ≈ 22.99 g/mol Atomic mass of Cl ≈ 35.45 g/mol
Molar mass of NaCl = Atomic mass of Na + Atomic mass of Cl Molar mass of NaCl = 22.99 g/mol + 35.45 g/mol Molar mass of NaCl = 58.44 g/mol
This means that one mole of NaCl has a mass of 58.44 grams. To convert moles to grams, we use the formula:
Mass = Moles × Molar Mass
We have 0.0109 moles of NaCl and the molar mass is 58.44 g/mol. Plugging these values into the formula:
Mass of NaCl = 0.0109 moles × 58.44 g/mol Mass of NaCl ≈ 0.637 g
Therefore, approximately 0.637 grams of NaCl are formed when 0.25 g of Na reacts completely with 0.39 g of Cl₂. Rounding this value to two significant figures, we get 0.64 g of NaCl. This is the theoretical yield of NaCl in this reaction. Converting moles to grams is a fundamental skill in chemistry, allowing us to relate the amount of substance in moles to its mass in grams. This conversion is essential for various applications, such as preparing solutions of specific concentrations, determining the yield of a reaction, and performing quantitative analysis. By understanding how to convert between moles and grams, we can accurately measure and manipulate chemical substances in the laboratory and in industrial processes. In conclusion, the mass of NaCl formed when 0.25 g of Na reacts completely with 0.39 g of Cl₂ is approximately 0.64 g. This calculation involves several key steps, including balancing the chemical equation, converting grams to moles, identifying the limiting reactant, calculating moles of product, and converting moles of product to grams. By mastering these steps, we can confidently solve a wide range of stoichiometry problems and gain a deeper understanding of chemical reactions.
Final Answer
Based on our calculations, the mass of NaCl formed when 0.25 g of Na reacts completely with 0.39 g of Cl₂ is approximately 0.64 g. Therefore, the correct answer is:
B. 0.64 g NaCl
What is the mass of NaCl produced when 0.25 g of Na reacts completely with 0.39 g of Cl₂?
NaCl Formation Calculation How Much NaCl from 0.25g Na and 0.39g Cl2
