Grams Of Oxygen Needed React Completely With NO Stoichiometry Calculation

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Introduction

In chemistry, understanding stoichiometry is crucial for predicting the amounts of reactants and products involved in a chemical reaction. Stoichiometry is the calculation of quantitative, or measurable, relationships of the reactants and products in balanced chemical equations. In this article, we will delve into a practical stoichiometry problem, focusing on calculating the mass of oxygen ($O_2$) required to react completely with a given mass of nitric oxide (NO). We will utilize the balanced chemical equation $2 NO + O_2 ightarrow 2 NO_2$ as our framework, and meticulously walk through each step of the calculation. This includes converting grams of NO to moles, using the stoichiometric ratio from the balanced equation to find moles of $O_2$ required, and finally, converting moles of $O_2$ back to grams. We will also ensure that our final answer adheres to the rules of significant figures, a fundamental aspect of accurate scientific reporting. By understanding these principles, you can predict the outcome of chemical reactions and effectively conduct experiments in the laboratory. Mastering stoichiometric calculations is a foundational skill that will serve you well in more advanced chemistry topics and practical applications. Let's embark on this journey to master stoichiometry and enhance our problem-solving abilities in chemistry. The importance of stoichiometry extends beyond the classroom, playing a crucial role in various fields such as pharmaceuticals, manufacturing, and environmental science. Accurate stoichiometric calculations are essential for ensuring the correct proportions of reactants in drug synthesis, optimizing industrial chemical processes, and assessing the environmental impact of chemical reactions. Therefore, a solid understanding of stoichiometry is not only academically beneficial but also practically relevant in numerous real-world scenarios.

Problem Statement

The question we aim to address is: How many grams of $O_2$ are needed to react completely with 23.1 g of NO, given the balanced equation $2 NO + O_2 ightarrow 2 NO_2$? It is crucial to ensure that the final answer has the correct number of significant figures. To solve this stoichiometry problem, we must first convert the given mass of NO into moles using its molar mass. The molar mass of a compound is the sum of the atomic masses of each element in the compound, expressed in grams per mole (g/mol). For NO, nitrogen (N) has an atomic mass of approximately 14.01 g/mol, and oxygen (O) has an atomic mass of approximately 16.00 g/mol. Therefore, the molar mass of NO is 14.01 g/mol + 16.00 g/mol = 30.01 g/mol. Once we have the number of moles of NO, we will use the balanced chemical equation to determine the mole ratio between NO and $O_2$. This ratio is essential for calculating the number of moles of $O_2$ required to react completely with the given amount of NO. The balanced equation $2 NO + O_2 ightarrow 2 NO_2$ tells us that 2 moles of NO react with 1 mole of $O_2$, providing us with the stoichiometric ratio needed for our calculation. Finally, we will convert the moles of $O_2$ back to grams using the molar mass of $O_2$. The molar mass of $O_2$ is 2 times the atomic mass of oxygen, which is 2 * 16.00 g/mol = 32.00 g/mol. By following these steps carefully, we can accurately determine the mass of $O_2$ required for the complete reaction. Ensuring the correct number of significant figures in our final answer is vital for maintaining scientific accuracy and precision. Significant figures represent the digits in a number that are known with certainty plus one uncertain digit, and they play a crucial role in quantitative analysis.

Step-by-Step Solution

Step 1: Convert grams of NO to moles of NO

To begin, we need to convert the given mass of NO (23.1 g) into moles. To do this, we use the molar mass of NO, which, as we calculated earlier, is 30.01 g/mol. The conversion formula is:

Moles of NO=Mass of NOMolar mass of NO\text{Moles of NO} = \frac{\text{Mass of NO}}{\text{Molar mass of NO}}

Plugging in the values:

Moles of NO=23.1 g30.01 g/mol=0.7697 mol\text{Moles of NO} = \frac{23.1 \text{ g}}{30.01 \text{ g/mol}} = 0.7697 \text{ mol}

So, 23.1 g of NO is equivalent to approximately 0.7697 moles. It's important to retain extra digits during intermediate calculations to avoid rounding errors, which can affect the accuracy of the final result. In this step, we have successfully converted the mass of NO into moles, which is a crucial step in stoichiometric calculations. The ability to convert between mass and moles is fundamental in chemistry, as it allows us to relate the macroscopic world (grams) to the microscopic world (moles of molecules). This conversion factor, the molar mass, acts as a bridge between these two scales, enabling us to perform quantitative analysis of chemical reactions. As we proceed through the problem, we will see how this initial conversion lays the groundwork for determining the amount of oxygen required for the reaction. The precision of this calculation is paramount, as it directly impacts the accuracy of subsequent steps. By using the molar mass and applying the correct formula, we ensure that our starting point is solid, setting the stage for a reliable final answer. Mastering this conversion is not just about solving this specific problem; it's about building a foundational skill that will be invaluable in various chemical calculations and applications.

Step 2: Use the balanced equation to find the mole ratio of $O_2$ to NO

The balanced chemical equation, $2 NO + O_2 ightarrow 2 NO_2$, provides the crucial information about the stoichiometric relationship between NO and $O_2$. From the equation, we can see that 2 moles of NO react with 1 mole of $O_2$. This gives us the mole ratio:

Moles of O2Moles of NO=12\frac{\text{Moles of } O_2}{\text{Moles of NO}} = \frac{1}{2}

This ratio is the key to determining how many moles of $O_2$ are needed to react completely with 0.7697 moles of NO. By understanding and applying mole ratios, we can predict the amounts of reactants and products involved in a chemical reaction. This is one of the fundamental principles of stoichiometry, allowing us to move from a qualitative understanding of a reaction (what reacts with what) to a quantitative understanding (how much of each substance is involved). The balanced chemical equation serves as a recipe, telling us the exact proportions of reactants and products. Without a balanced equation, we cannot accurately determine these proportions, and our calculations would be meaningless. The coefficients in the balanced equation represent the number of moles of each substance involved, and these coefficients are the foundation for establishing mole ratios. In this step, we are essentially translating the information encoded in the balanced equation into a practical tool for our calculations. The mole ratio is a dimensionless quantity, meaning it has no units, which makes it a versatile tool for converting between moles of different substances. As we move forward, this mole ratio will allow us to bridge the gap between the moles of NO we calculated in the previous step and the moles of $O_2$ we need to find.

Step 3: Calculate moles of $O_2$ needed

Now that we have the mole ratio of $O_2$ to NO (1/2), we can calculate the number of moles of $O_2$ required to react completely with 0.7697 moles of NO. Using the mole ratio, we set up the equation:

Moles of O2=Moles of NO×Moles of O2Moles of NO\text{Moles of } O_2 = \text{Moles of NO} \times \frac{\text{Moles of } O_2}{\text{Moles of NO}}

Plugging in the values:

Moles of O2=0.7697 mol×12=0.38485 mol\text{Moles of } O_2 = 0.7697 \text{ mol} \times \frac{1}{2} = 0.38485 \text{ mol}

Therefore, 0.38485 moles of $O_2$ are needed to react completely with 23.1 g of NO. This step demonstrates the power of stoichiometry in predicting the quantities of substances involved in a chemical reaction. By applying the mole ratio derived from the balanced equation, we have successfully determined the amount of $O_2$ required to react with a specific amount of NO. This calculation is a cornerstone of chemical synthesis and analysis, allowing chemists to precisely control the outcome of reactions. The ability to calculate the required amounts of reactants is essential for optimizing reaction yields and minimizing waste. In this context, understanding mole ratios enables us to scale reactions up or down, ensuring that the correct proportions of reactants are used regardless of the scale. This is particularly important in industrial chemistry, where large-scale reactions are common. Furthermore, this step highlights the interconnectedness of the previous steps. The accuracy of the moles of $O_2$ calculation depends directly on the accuracy of the moles of NO calculation in Step 1 and the correct interpretation of the mole ratio in Step 2. This underscores the importance of a methodical and precise approach to stoichiometric problems. As we move to the final step, we will convert these moles of $O_2$ into grams, providing us with the answer in the desired unit.

Step 4: Convert moles of $O_2$ to grams of $O_2$

To find the mass of $O_2$ needed, we convert moles of $O_2$ (0.38485 mol) to grams using the molar mass of $O_2$. The molar mass of $O_2$ is 32.00 g/mol. The conversion formula is:

Grams of O2=Moles of O2×Molar mass of O2\text{Grams of } O_2 = \text{Moles of } O_2 \times \text{Molar mass of } O_2

Plugging in the values:

Grams of O2=0.38485 mol×32.00 g/mol=12.3152 g\text{Grams of } O_2 = 0.38485 \text{ mol} \times 32.00 \text{ g/mol} = 12.3152 \text{ g}

So, we have calculated that 12.3152 grams of $O_2$ are needed to react completely with 23.1 g of NO. This step completes the stoichiometric calculation, providing us with the mass of oxygen required for the reaction. Converting moles to grams is a fundamental skill in chemistry, as it allows us to express the amount of a substance in a unit that is easily measurable in the laboratory. The molar mass acts as the conversion factor, bridging the gap between the mole, a unit of amount, and the gram, a unit of mass. This conversion is essential for practical applications, such as weighing out reactants for a chemical reaction or determining the yield of a product. In this context, the accurate determination of the mass of $O_2$ required is crucial for ensuring that the reaction proceeds as expected. An insufficient amount of $O_2$ would result in incomplete reaction, while an excessive amount might lead to unwanted side reactions. Therefore, this step highlights the importance of precision in stoichiometric calculations. As we move to the final part of the solution, we will consider the significant figures to ensure that our answer reflects the precision of the given data.

Step 5: Apply significant figures

The final step is to apply the rules of significant figures to our answer. In the original problem, we were given 23.1 g of NO, which has three significant figures. The molar mass of $O_2$ (32.00 g/mol) has four significant figures, and the molar mass of NO (30.01 g/mol) has four significant figures. In multiplication and division, the result should have the same number of significant figures as the measurement with the fewest significant figures.

Our calculated mass of $O_2$ is 12.3152 g. Since 23.1 g of NO has three significant figures, our final answer should also have three significant figures. Rounding 12.3152 g to three significant figures gives us 12.3 g.

Therefore, 12.3 grams of $O_2$ are needed to react completely with 23.1 g of NO. Applying the rules of significant figures is crucial in scientific calculations to accurately represent the precision of the measurements. Significant figures indicate the digits in a number that are known with certainty plus one doubtful digit. They are a way of conveying the uncertainty associated with a measurement. In this context, rounding our final answer to three significant figures ensures that we are not overstating the precision of our result. The number of significant figures in the given data limits the precision of our final answer. For example, if we had been given 23.10 g of NO (four significant figures), our final answer could have four significant figures as well. This step highlights the importance of considering the limitations of our measurements and reporting our results accordingly. By adhering to the rules of significant figures, we maintain scientific integrity and ensure that our results are meaningful and reliable. In summary, this final step is a critical aspect of problem-solving in chemistry, ensuring that our calculated result is both accurate and appropriately precise.

Final Answer

Therefore, 12.3 grams of $O_2$ are needed to react completely with 23.1 g of NO, ensuring the answer has the correct number of significant figures.

Conclusion

In this comprehensive walkthrough, we have successfully determined the amount of oxygen required to react completely with a given amount of nitric oxide, adhering to the principles of stoichiometry and significant figures. We began by converting the mass of NO to moles, utilizing the molar mass of NO. This conversion allowed us to work with the stoichiometric relationships defined by the balanced chemical equation. Next, we employed the balanced equation to establish the mole ratio between NO and $O_2$, a crucial step in determining the required amount of $O_2$. Using this ratio, we calculated the moles of $O_2$ needed for the reaction. Following this, we converted the moles of $O_2$ back to grams, providing us with the mass of oxygen required. Finally, we meticulously applied the rules of significant figures to ensure our answer reflected the precision of the initial measurements. The final result, 12.3 grams of $O_2$, represents the mass of oxygen required for the complete reaction. This exercise underscores the importance of a systematic approach to solving stoichiometric problems. Each step builds upon the previous one, and accuracy at each stage is paramount for achieving a reliable final result. By mastering these steps, you can confidently tackle a wide range of stoichiometry problems, which are fundamental to understanding chemical reactions and quantitative analysis in chemistry. Furthermore, this detailed solution highlights the practical application of chemical principles. Stoichiometry is not merely a theoretical concept; it is a tool that chemists use every day to predict and control the outcomes of chemical reactions. Whether in the laboratory, the industrial setting, or environmental research, the ability to perform stoichiometric calculations accurately is essential for success. Therefore, a solid understanding of stoichiometry is an invaluable asset for anyone pursuing a career in chemistry or a related field.