Calculating MnO2 Mass From H2O Reaction Stoichiometry In Chemistry

In the realm of chemistry, understanding stoichiometry and redox reactions is paramount for predicting the quantities of reactants and products involved in a chemical transformation. This article delves into a specific chemical reaction, exploring the process of calculating the mass of manganese dioxide ($MnO_2$) produced when a given mass of water ($H_2O$) reacts with potassium permanganate ($MnO_4^−$) and bromide ions ($Br^−$). This type of calculation is crucial in various chemical applications, from laboratory experiments to industrial processes, where precise control over reaction outcomes is essential. To achieve an accurate result, we will meticulously analyze the balanced chemical equation, apply stoichiometric principles, and perform the necessary molar mass conversions. Mastering these techniques is vital for anyone seeking a comprehensive understanding of quantitative chemical analysis. This article provides a step-by-step guide to solving this specific problem, emphasizing the importance of understanding the underlying principles and methodologies involved in stoichiometry. By following this guide, readers will not only be able to solve this particular problem but also gain the skills and knowledge necessary to tackle similar stoichiometric calculations in various chemical contexts. Understanding the relationship between reactants and products in a chemical reaction is fundamental to predicting the outcome and efficiency of chemical processes, making this a core concept in chemistry education and practice.

The chemical reaction in question involves the reaction of water ($H_2O$) with permanganate ions ($MnO_4^−$) and bromide ions ($Br^−$) to produce bromate ions ($BrO_3^−$), manganese dioxide ($MnO_2$), and hydroxide ions ($OH^−$). The balanced chemical equation for this reaction is:

$H_2O + 2MnO_4^− + Br^− → BrO_3^− + 2MnO_2 + 2OH^−$

This equation is essential as it provides the stoichiometric ratios between the reactants and products. Stoichiometry, the quantitative relationship between reactants and products in a chemical reaction, is governed by the law of conservation of mass, which states that matter cannot be created or destroyed in a chemical reaction. This law is reflected in the balanced chemical equation, where the number of atoms of each element is the same on both sides of the equation. In this specific reaction, the equation shows that one mole of water ($H_2O$) reacts with two moles of permanganate ions ($MnO_4^−$) and one mole of bromide ions ($Br^−$) to produce one mole of bromate ions ($BrO_3^−$), two moles of manganese dioxide ($MnO_2$), and two moles of hydroxide ions ($OH^−$). Understanding these molar ratios is crucial for determining the amount of $MnO_2$ produced from a given amount of $H_2O$. For instance, the equation clearly indicates that for every one mole of $H_2O$ consumed, two moles of $MnO_2$ are produced. This 1:2 molar ratio between $H_2O$ and $MnO_2$ is the key to solving the problem. By converting the given mass of $H_2O$ to moles, we can use this ratio to calculate the moles of $MnO_2$ produced and subsequently convert this back to mass. The balanced chemical equation not only provides the molar ratios but also ensures that the overall charge and the number of atoms of each element are conserved throughout the reaction. This balance is a fundamental principle in chemistry, ensuring that the equation accurately represents the chemical transformation taking place.

The problem asks us to determine the mass of $MnO_2$ produced when 445 grams of $H_2O$ react. To solve this problem, we will follow a series of steps based on stoichiometric principles. These steps involve converting the mass of $H_2O$ to moles, using the balanced chemical equation to find the moles of $MnO_2$ produced, and finally, converting the moles of $MnO_2$ back to grams. This type of calculation is a fundamental aspect of quantitative chemistry, where the ability to predict the amount of product formed from a given amount of reactant is crucial. The problem highlights the practical application of stoichiometry, a branch of chemistry that deals with the quantitative relationships between reactants and products in chemical reactions. By understanding stoichiometry, chemists can predict the amount of product formed in a reaction, determine the limiting reactant, and optimize reaction conditions for maximum yield. In this specific problem, the mass of $H_2O$ is given, and we need to find the corresponding mass of $MnO_2$ produced. This requires a clear understanding of molar masses and the molar ratio between $H_2O$ and $MnO_2$ as dictated by the balanced chemical equation. The problem emphasizes the importance of accurate conversions between grams and moles, as well as the correct application of the stoichiometric coefficients from the balanced equation. By working through this problem, we reinforce the fundamental skills necessary for solving a wide range of quantitative chemistry problems.

Step 1: Convert grams of $H_2O$ to moles

First, we need to convert the mass of $H_2O$ given (445 grams) to moles. To do this, we use the molar mass of $H_2O$, which is approximately 18.015 g/mol.

Moles of $H_2O$ = (Mass of $H_2O$) / (Molar mass of $H_2O$)

Moles of $H_2O$ = 445 g / 18.015 g/mol

Moles of $H_2O$ ≈ 24.70 mol

This conversion is a crucial first step because stoichiometric calculations are based on molar ratios, not mass ratios. The molar mass of a substance is the mass of one mole of that substance, and it serves as the bridge between mass and moles. In this case, the molar mass of water ($H_2O$) is the sum of the atomic masses of its constituent atoms: two hydrogen atoms (approximately 1.008 g/mol each) and one oxygen atom (approximately 16.00 g/mol). Therefore, the molar mass of $H_2O$ is approximately 2(1.008) + 16.00 = 18.016 g/mol. Using this molar mass, we can convert the given mass of water (445 grams) to moles by dividing the mass by the molar mass. This gives us the amount of water in moles, which is essential for using the stoichiometric ratios from the balanced chemical equation. The accurate determination of molar masses is fundamental in chemistry, as it allows for the precise calculation of amounts of substances in chemical reactions. Failing to use the correct molar mass would lead to significant errors in subsequent calculations, highlighting the importance of this initial conversion step.

Step 2: Use the stoichiometric ratio to find moles of $MnO_2$

From the balanced equation, we see that 1 mole of $H_2O$ produces 2 moles of $MnO_2$. Therefore, the mole ratio of $H_2O$ to $MnO_2$ is 1:2.

Moles of $MnO_2$ = Moles of $H_2O$ × (2 moles $MnO_2$ / 1 mole $H_2O$)

Moles of $MnO_2$ = 24.70 mol × 2

Moles of $MnO_2$ ≈ 49.40 mol

This step is the heart of stoichiometric calculations, where the balanced chemical equation is used to relate the amounts of different substances involved in the reaction. The coefficients in the balanced equation represent the molar ratios of the reactants and products. In this case, the balanced equation $H_2O + 2MnO_4^− + Br^− → BrO_3^− + 2MnO_2 + 2OH^−$ tells us that for every 1 mole of water ($H_2O$) that reacts, 2 moles of manganese dioxide ($MnO_2$) are produced. This 1:2 molar ratio is the key to converting the moles of $H_2O$ calculated in the previous step to moles of $MnO_2$. To find the moles of $MnO_2$ produced, we multiply the moles of $H_2O$ by this ratio. This step directly applies the law of definite proportions, which states that a chemical compound always contains exactly the same proportion of elements by mass. In other words, the ratio of moles of reactants and products in a chemical reaction is fixed, as defined by the balanced equation. Understanding and correctly applying these stoichiometric ratios is crucial for accurate quantitative analysis in chemistry. Errors in this step can propagate through the rest of the calculation, leading to incorrect results. Therefore, careful attention to the balanced equation and the molar ratios it provides is essential for solving stoichiometry problems.

Step 3: Convert moles of $MnO_2$ to grams

Now, we convert the moles of $MnO_2$ to grams using its molar mass. The molar mass of $MnO_2$ is approximately 86.94 g/mol.

Mass of $MnO_2$ = Moles of $MnO_2$ × (Molar mass of $MnO_2$)

Mass of $MnO_2$ = 49.40 mol × 86.94 g/mol

Mass of $MnO_2$ ≈ 4295.84 g

Rounding this to the nearest hundred gives us approximately 4,300 g.

This final conversion step brings us back to the units we were originally asked for: grams of $MnO_2$. To convert moles of $MnO_2$ to grams, we use the molar mass of $MnO_2$, which is the mass of one mole of $MnO_2$. The molar mass is calculated by summing the atomic masses of the elements in the compound, which are obtained from the periodic table. For $MnO_2$, this is the sum of the atomic mass of manganese (Mn) and twice the atomic mass of oxygen (O). The atomic mass of Mn is approximately 54.94 g/mol, and the atomic mass of O is approximately 16.00 g/mol. Therefore, the molar mass of $MnO_2$ is approximately 54.94 + 2(16.00) = 86.94 g/mol. Multiplying the moles of $MnO_2$ (49.40 mol) by its molar mass (86.94 g/mol) gives us the mass of $MnO_2$ produced in grams. This step completes the stoichiometric calculation, providing the answer to the original problem. The final rounding is done to match the precision of the given options, ensuring that the answer is presented in a clear and meaningful way. This entire process, from converting grams of $H_2O$ to moles, using the stoichiometric ratio to find moles of $MnO_2$, and then converting moles of $MnO_2$ back to grams, demonstrates the power and utility of stoichiometry in solving quantitative chemical problems. Understanding these steps is crucial for anyone working in chemistry, from students to researchers and industrial chemists.

The mass of $MnO_2$ produced is approximately 4,300 g.

Therefore, the correct answer is:

B. 4,300 g

In conclusion, determining the mass of $MnO_2$ produced from the reaction of 445 grams of $H_2O$ involves a series of stoichiometric calculations. By converting the mass of $H_2O$ to moles, using the balanced chemical equation to find the mole ratio between $H_2O$ and $MnO_2$, and finally, converting the moles of $MnO_2$ back to grams, we arrived at the answer of approximately 4,300 g. This process highlights the importance of understanding stoichiometry and molar mass conversions in quantitative chemical analysis. Mastering these skills is essential for solving a wide range of chemistry problems and for making accurate predictions in chemical reactions. The steps involved in this calculation are not unique to this specific problem but are applicable to many other stoichiometric calculations in chemistry. Understanding the relationship between reactants and products in a chemical reaction is fundamental to predicting the outcome and efficiency of chemical processes. The ability to accurately calculate the amount of product formed from a given amount of reactant is a critical skill in various fields, including chemical research, industrial chemistry, and environmental science. Furthermore, this exercise reinforces the importance of paying close attention to units and significant figures in chemical calculations. Consistent use of units and proper rounding techniques are crucial for obtaining accurate and reliable results. By working through this problem, we have not only found the answer but also strengthened our understanding of the underlying chemical principles and techniques. This understanding is the foundation for more advanced topics in chemistry and is invaluable for anyone pursuing a career in the chemical sciences.