Stoichiometry Calculation Grams Of NaOH React With MgCl2

JU07/30/2025 08, 2025 by THE IDEN 57 views

Stoichiometry Calculation Sodium Hydroxide and Magnesium Chloride Reaction

Introduction to Stoichiometry in Chemical Reactions

In the realm of chemistry, understanding stoichiometry is paramount for predicting and quantifying the outcomes of chemical reactions. Stoichiometry, derived from the Greek words stoicheion (element) and metron (measure), essentially deals with the quantitative relationships between reactants and products in chemical reactions. It's the cornerstone for calculating the amounts of substances consumed and produced in chemical processes, ensuring accuracy and efficiency in various applications, from industrial chemical synthesis to laboratory experiments. This article delves into a practical stoichiometry problem, focusing on the reaction between sodium hydroxide (NaOH) and magnesium chloride (MgCl₂) to produce magnesium hydroxide (Mg(OH)₂) and sodium chloride (NaCl). By meticulously examining the balanced chemical equation and employing stoichiometric principles, we will unravel the quantitative aspects of this reaction, providing a clear understanding of the molar relationships and mass transformations involved.

Stoichiometry calculations are essential for several reasons. Firstly, they allow chemists to optimize reactions by ensuring the correct amounts of reactants are used, maximizing product yield and minimizing waste. Secondly, stoichiometric calculations are crucial in industrial processes where large-scale chemical reactions need to be precisely controlled for economic and safety reasons. Thirdly, in research settings, stoichiometry helps in designing experiments, interpreting results, and understanding the underlying mechanisms of chemical reactions. A solid grasp of stoichiometry empowers students and professionals alike to navigate the intricacies of chemical transformations with confidence and precision. Moreover, the ability to perform stoichiometric calculations is a fundamental skill tested in chemistry courses and professional examinations, highlighting its significance in the broader scientific community. In the subsequent sections, we will apply these principles to solve a specific problem, reinforcing the practical application of stoichiometry in a chemical context.

Problem Statement: Reaction of Sodium Hydroxide and Magnesium Chloride

The problem we're addressing involves the reaction between sodium hydroxide (NaOH) and magnesium chloride (MgCl₂). The balanced chemical equation for this reaction is:

$2 NaOH + MgCl_2 \rightarrow Mg(OH)_2 + 2 NaCl$

This equation tells us that two moles of sodium hydroxide react with one mole of magnesium chloride to produce one mole of magnesium hydroxide and two moles of sodium chloride. Stoichiometrically balanced equations like this are the foundation for all quantitative calculations in chemistry, as they provide the precise molar ratios between reactants and products.

The specific scenario we're examining begins with 637 grams of magnesium chloride (MgCl₂). Our goal is to determine the mass of sodium hydroxide (NaOH) required to completely react with this amount of magnesium chloride. This is a classic stoichiometry problem that requires us to convert mass to moles, apply the stoichiometric ratio from the balanced equation, and then convert back to mass. This step-by-step approach ensures we accurately determine the quantity of sodium hydroxide needed for the reaction. Solving this problem will not only demonstrate the practical application of stoichiometry but also reinforce the importance of using molar masses and stoichiometric ratios in chemical calculations.

To solve this problem effectively, we need to follow a systematic approach. First, we will calculate the number of moles of magnesium chloride present using its molar mass. Next, we will use the stoichiometric ratio from the balanced equation to determine the number of moles of sodium hydroxide required. Finally, we will convert the moles of sodium hydroxide back to grams using its molar mass. By breaking down the problem into these manageable steps, we can ensure accuracy and clarity in our calculations. The ability to perform such calculations is vital for anyone working in chemistry, whether in a laboratory, an industrial setting, or an academic environment. In the following sections, we will delve into the detailed calculations, showing how each step is performed and why it is necessary for arriving at the correct answer.

Step-by-Step Solution: Determining the Mass of Sodium Hydroxide

Step 1: Calculate the Moles of Magnesium Chloride ( $MgCl_2$ )

The first step in solving this stoichiometry problem is to determine the number of moles of magnesium chloride ( $MgCl_2$ ) present. To do this, we need to use the molar mass of $MgCl_2$ . The molar mass is the mass of one mole of a substance, and it's calculated by summing the atomic masses of all the atoms in the chemical formula. For $MgCl_2$ , the molar mass is calculated as follows:

Molar mass of Mg = 24.31 g/mol
Molar mass of Cl = 35.45 g/mol

Therefore, the molar mass of $MgCl_2$ is:

$24.31 rac{g}{mol} + 2 imes 35.45 rac{g}{mol} = 95.21 rac{g}{mol}$

Now that we have the molar mass of $MgCl_2$ , we can calculate the number of moles using the formula:

$Moles = rac{Mass}{Molar\ Mass}$

Given that we have 637 grams of $MgCl_2$ , the number of moles is:

$Moles\ of\ MgCl_2 = rac{637\ g}{95.21\ g/mol} = 6.69\ mol$ (rounded to three significant figures)

This calculation is a crucial first step because it converts the given mass of $MgCl_2$ into a quantity that we can use in the stoichiometric ratios from the balanced chemical equation. The ability to accurately convert between mass and moles is a fundamental skill in chemistry, and it is essential for solving a wide range of quantitative problems. In the next step, we will use this value, along with the balanced chemical equation, to determine the moles of NaOH required for the reaction. Understanding the molar relationship between reactants and products is the key to mastering stoichiometry, and this step sets the stage for the subsequent calculations. By performing this conversion accurately, we ensure that our subsequent calculations are based on the correct molar quantities, leading to a precise final answer.

Step 2: Determine the Moles of Sodium Hydroxide (NaOH) Required

Having calculated the moles of magnesium chloride ( $MgCl_2$ ), we can now determine the number of moles of sodium hydroxide (NaOH) required for the reaction. This step utilizes the stoichiometric ratio from the balanced chemical equation:

$2 NaOH + MgCl_2 \rightarrow Mg(OH)_2 + 2 NaCl$

From the equation, we can see that 2 moles of NaOH react with 1 mole of $MgCl_2$ . This gives us the stoichiometric ratio:

$rac{Moles\ of\ NaOH}{Moles\ of\ MgCl_2} = rac{2}{1}$

Using this ratio, we can calculate the moles of NaOH required to react with 6.69 moles of $MgCl_2$ :

$Moles\ of\ NaOH = 2 imes Moles\ of\ MgCl_2$ $Moles\ of\ NaOH = 2 imes 6.69\ mol$ $Moles\ of\ NaOH = 13.38\ mol$ (rounded to three significant figures)

This calculation is a direct application of the stoichiometric coefficients from the balanced equation. These coefficients provide the molar relationships between reactants and products, allowing us to determine the exact amount of one substance needed to react with a given amount of another. The stoichiometric ratio acts as a conversion factor, enabling us to move from the moles of one substance to the moles of another. This step is critical in stoichiometry because it bridges the gap between the known quantity of one reactant and the required quantity of another. Accurate determination of these molar relationships is essential for optimizing reactions and predicting product yields. In the final step, we will convert these moles of NaOH back into grams to provide the final answer in the desired units. The ability to use stoichiometric ratios correctly is a fundamental skill in chemistry, and this step exemplifies its importance in solving quantitative problems.

Step 3: Convert Moles of Sodium Hydroxide to Grams

The final step in our stoichiometry calculation is to convert the moles of sodium hydroxide (NaOH) required into grams. To do this, we will use the molar mass of NaOH. The molar mass is calculated by summing the atomic masses of each element in the chemical formula:

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

Therefore, the molar mass of NaOH is:

$22.99 rac{g}{mol} + 16.00 rac{g}{mol} + 1.01 rac{g}{mol} = 40.00 rac{g}{mol}$

Now that we have the molar mass of NaOH, we can convert the moles of NaOH to grams using the formula:

$Mass = Moles imes Molar\ Mass$

We calculated that 13.38 moles of NaOH are required, so the mass of NaOH needed is:

$Mass\ of\ NaOH = 13.38\ mol imes 40.00 rac{g}{mol}$ $Mass\ of\ NaOH = 535.2\ g$

Rounding this to three significant figures, we get:

$Mass\ of\ NaOH = 535\ g$

This final conversion completes our stoichiometry problem, providing the answer in the desired units of grams. Converting moles to grams is a common step in stoichiometry calculations, as it allows us to relate the molar quantities to the actual masses of substances used in the laboratory or industrial setting. This step highlights the practical application of stoichiometry, enabling chemists to accurately measure out the required amounts of reactants for a chemical reaction. The molar mass acts as a conversion factor, linking the microscopic world of moles to the macroscopic world of grams. By performing this calculation accurately, we ensure that we have the correct amount of NaOH to react completely with the 637 grams of $MgCl_2$ . This entire process demonstrates the power of stoichiometry in quantitatively analyzing chemical reactions and predicting the amounts of substances involved.

Final Answer and Conclusion

Final Answer: Mass of Sodium Hydroxide Required

After performing the stoichiometry calculations, we have determined the mass of sodium hydroxide (NaOH) required to react completely with 637 grams of magnesium chloride ( $MgCl_2$ ). Our calculations have shown that:

535 grams of NaOH are required to react completely with 637 grams of $MgCl_2$ .

This final answer, expressed to three significant figures, provides a precise quantity of NaOH needed for the reaction. The process involved converting the mass of $MgCl_2$ to moles, using the stoichiometric ratio from the balanced chemical equation to find the moles of NaOH required, and then converting those moles back into grams. This step-by-step approach is a hallmark of stoichiometry problem-solving, ensuring accuracy and clarity in the calculations. This answer is crucial for anyone planning to carry out this reaction, whether in a laboratory setting or an industrial process, as it ensures the correct amounts of reactants are used, maximizing the yield of products and minimizing waste. The ability to perform these calculations accurately is a testament to a solid understanding of stoichiometric principles and their application in real-world scenarios.

Conclusion: Significance of Stoichiometry in Chemistry

In conclusion, this exercise in calculating the mass of sodium hydroxide required to react with magnesium chloride highlights the fundamental importance of stoichiometry in chemistry. Stoichiometry is not just a mathematical exercise; it is a critical tool for understanding and quantifying chemical reactions. By using balanced chemical equations and molar masses, we can accurately predict the amounts of reactants needed and products formed in a chemical reaction. This ability is essential for a wide range of applications, from research and development to industrial chemical processes.

Stoichiometry allows chemists to:

Optimize reactions: By using the correct amounts of reactants, we can maximize product yield and minimize waste.
Scale-up reactions: Stoichiometric calculations are crucial for scaling up reactions from the laboratory to industrial production, ensuring that the process remains efficient and safe.
Analyze experimental data: Stoichiometry helps in interpreting experimental results and determining the efficiency of a reaction.
Design new chemical processes: By understanding the quantitative relationships between reactants and products, chemists can design new chemical processes and reactions.

The problem-solving approach demonstrated in this article, involving conversion of mass to moles, application of stoichiometric ratios, and conversion back to mass, is a standard technique in chemistry. Mastering these steps is essential for students and professionals alike. A strong grasp of stoichiometry enables chemists to navigate the complexities of chemical reactions with confidence and precision. This example underscores the practical significance of stoichiometry and its indispensable role in the field of chemistry.