Calculate Moles Of Hydrogen Gas Produced From 138 G NaOH

Introduction: Unveiling the Stoichiometry of Sodium Hydroxide and Hydrogen Gas Production

In the realm of chemistry, understanding the quantitative relationships between reactants and products in chemical reactions is paramount. This concept, known as stoichiometry, allows us to predict the amounts of substances consumed and produced in a chemical reaction. In this comprehensive exploration, we delve into the reaction between sodium (Na) and water (H₂O) to produce sodium hydroxide (NaOH) and hydrogen gas (H₂), specifically focusing on determining the number of moles of hydrogen gas (H₂) generated when 138 grams of sodium hydroxide (NaOH) are produced. This analysis involves applying stoichiometric principles and molar mass calculations to unravel the quantitative connections within this chemical transformation.

To address this intriguing question, we will meticulously examine the balanced chemical equation for the reaction, which serves as a cornerstone for stoichiometric calculations. The balanced equation reveals the precise molar ratios between reactants and products, providing the essential framework for our analysis. Subsequently, we will embark on a step-by-step journey, converting the given mass of sodium hydroxide (NaOH) into moles, leveraging its molar mass as a conversion factor. Once we have determined the moles of NaOH produced, we will employ the stoichiometric ratio derived from the balanced equation to calculate the corresponding moles of hydrogen gas (H₂) generated concurrently. This meticulous approach will enable us to accurately quantify the amount of hydrogen gas produced in conjunction with the specified amount of sodium hydroxide, elucidating the stoichiometric relationship between these two chemical species.

This exploration will not only provide a concrete answer to the question at hand but also serve as a valuable exercise in applying stoichiometric principles, which are fundamental to understanding and predicting chemical reactions. By mastering these concepts, we gain the ability to navigate the quantitative aspects of chemistry, paving the way for deeper insights into the behavior of chemical substances and their interactions.

Decoding the Chemical Equation: The Foundation of Stoichiometric Calculations

The cornerstone of any stoichiometric calculation lies in the balanced chemical equation. This equation serves as a roadmap, delineating the exact molar ratios between reactants and products involved in a chemical reaction. For the reaction between sodium (Na) and water (H₂O) to produce sodium hydroxide (NaOH) and hydrogen gas (H₂), the balanced chemical equation is:

2 Na + 2 H₂O → 2 NaOH + H₂

This equation reveals a wealth of information, telling us that two moles of sodium (Na) react with two moles of water (H₂O) to produce two moles of sodium hydroxide (NaOH) and one mole of hydrogen gas (H₂). The coefficients in front of each chemical formula represent the stoichiometric coefficients, which are the key to unlocking the quantitative relationships within the reaction. In this specific case, the stoichiometric coefficient for NaOH is 2, while the coefficient for H₂ is 1. This crucial ratio signifies that for every two moles of NaOH produced, one mole of H₂ is generated simultaneously.

Understanding the stoichiometric ratios is paramount for accurate stoichiometric calculations. These ratios act as conversion factors, allowing us to translate between the amounts of different substances involved in the reaction. For instance, the ratio between NaOH and H₂ is 2:1, meaning that for every 2 moles of NaOH, 1 mole of H₂ is produced. This ratio will be instrumental in determining the moles of H₂ produced when 138 grams of NaOH are generated. By carefully examining the balanced chemical equation and extracting the relevant stoichiometric ratios, we lay the groundwork for precise calculations and a deeper understanding of the chemical transformation.

Step-by-Step Calculation: Unveiling the Moles of Hydrogen Gas

Now that we have the balanced chemical equation and a firm grasp of stoichiometric ratios, we can embark on the journey of calculating the moles of hydrogen gas (H₂) produced when 138 grams of sodium hydroxide (NaOH) are generated. This calculation involves a series of logical steps, each building upon the previous one, ultimately leading us to the desired answer.

Step 1: Converting Grams of NaOH to Moles

The first crucial step is to convert the given mass of NaOH (138 grams) into moles. To accomplish this, we need the molar mass of NaOH, which is the mass of one mole of NaOH. The molar mass of NaOH can be calculated by summing the atomic masses of its constituent elements (Na, O, and H) from the periodic table:

Molar mass of Na = 22.99 g/mol Molar mass of O = 16.00 g/mol Molar mass of H = 1.01 g/mol

Molar mass of NaOH = 22.99 + 16.00 + 1.01 = 39.99 g/mol

Now, we can use the molar mass as a conversion factor to convert grams of NaOH to moles:

Moles of NaOH = (Mass of NaOH) / (Molar mass of NaOH) Moles of NaOH = (138 g) / (39.99 g/mol) ≈ 3.45 moles

Step 2: Applying the Stoichiometric Ratio

With the moles of NaOH calculated, we can now leverage the stoichiometric ratio from the balanced chemical equation to determine the moles of H₂ produced. The balanced equation (2 Na + 2 H₂O → 2 NaOH + H₂) tells us that 2 moles of NaOH are produced for every 1 mole of H₂ generated. This 2:1 ratio is the key to our calculation.

Using this ratio, we can set up a proportion to find the moles of H₂:

(Moles of H₂) / (Moles of NaOH) = (1 mole H₂) / (2 moles NaOH)

Plugging in the moles of NaOH we calculated earlier (3.45 moles):

(Moles of H₂) / (3.45 moles) = 1/2

Solving for moles of H₂:

Moles of H₂ = (3.45 moles) / 2 ≈ 1.73 moles

Therefore, approximately 1.73 moles of hydrogen gas (H₂) will be produced at the same time as 138 grams of sodium hydroxide (NaOH).

Conclusion: Stoichiometry in Action - Unveiling Chemical Proportions

In this comprehensive exploration, we have successfully navigated the realm of stoichiometry to determine the moles of hydrogen gas (H₂) produced concurrently with 138 grams of sodium hydroxide (NaOH) in the reaction between sodium and water. By meticulously analyzing the balanced chemical equation, converting grams of NaOH to moles, and applying the stoichiometric ratio, we arrived at the answer of approximately 1.73 moles of H₂.

This exercise underscores the power of stoichiometry in predicting and quantifying the amounts of substances involved in chemical reactions. The balanced chemical equation serves as the foundation, providing the essential stoichiometric ratios that govern the relationships between reactants and products. By mastering these principles, we gain the ability to unravel the quantitative aspects of chemical transformations, enabling us to make accurate predictions and deepen our understanding of the chemical world.

The ability to perform stoichiometric calculations is not merely an academic exercise; it has profound implications in various fields, including chemical synthesis, industrial processes, and environmental science. In chemical synthesis, stoichiometry ensures that reactants are combined in the correct proportions to maximize product yield and minimize waste. In industrial processes, it plays a crucial role in optimizing reactions and ensuring efficient production. In environmental science, stoichiometry helps us understand and manage chemical reactions that impact our environment, such as the formation of pollutants or the degradation of materials.

In conclusion, our journey into the stoichiometry of the sodium hydroxide and hydrogen gas reaction has illuminated the importance of quantitative relationships in chemistry. By applying these principles, we can confidently predict and quantify the outcomes of chemical reactions, paving the way for advancements in various scientific and technological domains. The 1.73 moles of hydrogen gas produced alongside 138 grams of sodium hydroxide stand as a testament to the power of stoichiometry in unraveling the intricate proportions that govern the chemical world.

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