Sodium Chlorate Mass Calculation For 10.5 G Oxygen Production

In this article, we will walk through the process of calculating the mass of sodium chlorate ($NaClO_3$) needed to produce 10.5 g of oxygen ($O_2$), given the balanced chemical equation:

$2 NaClO_3(s) \rightarrow 2 NaCl(s) + 3 O_2(g)$

We are also given the molar masses of $O_2$ (32.00 g/mol) and $NaClO_3$ (106.44 g/mol). This calculation is a fundamental example of stoichiometry, which is the quantitative relationship between reactants and products in chemical reactions. Understanding stoichiometry is crucial in chemistry for predicting the amounts of substances consumed and produced in chemical reactions. Let's dive into the step-by-step solution.

Step 1: Convert the mass of oxygen to moles

The first step in any stoichiometry problem is to convert the given mass of a substance into moles. This conversion allows us to use the stoichiometric coefficients from the balanced chemical equation. To convert the mass of oxygen ($O_2$) to moles, we use the formula:

$Moles = \frac{Mass}{Molar\,Mass}$

In this case, we have 10.5 g of $O_2$, and the molar mass of $O_2$ is 32.00 g/mol. Plugging these values into the formula, we get:

$Moles\,of\,O_2 = \frac{10.5\,g}{32.00\,g/mol} = 0.328\,mol$

Therefore, 10.5 g of oxygen is equivalent to 0.328 moles. This conversion is crucial because the balanced chemical equation relates the number of moles of reactants and products, not their masses. The molar mass acts as a bridge between the macroscopic world of grams and the microscopic world of moles, allowing us to work with the fundamental units of chemical reactions.

Step 2: Use the stoichiometric ratio to find moles of sodium chlorate

The balanced chemical equation provides the stoichiometric ratios between the reactants and products. The equation:

$2 NaClO_3(s) \rightarrow 2 NaCl(s) + 3 O_2(g)$

shows that 2 moles of $NaClO_3$ decompose to produce 3 moles of $O_2$. This ratio is the key to finding out how many moles of $NaClO_3$ are needed to produce 0.328 moles of $O_2$. We can set up a proportion using the stoichiometric coefficients:

$\frac{Moles\,of\,NaClO_3}{Moles\,of\,O_2} = \frac{2}{3}$

Now, we can plug in the moles of $O_2$ we calculated in the previous step and solve for the moles of $NaClO_3$:

$\frac{Moles\,of\,NaClO_3}{0.328\,mol} = \frac{2}{3}$

$Moles\,of\,NaClO_3 = \frac{2}{3} \times 0.328\,mol = 0.219\,mol$

Thus, 0.219 moles of $NaClO_3$ are required to produce 0.328 moles of $O_2$. The stoichiometric ratio acts as a conversion factor, allowing us to move from the moles of one substance to the moles of another in a chemical reaction. This step is the heart of stoichiometry, as it directly applies the quantitative relationships described by the balanced equation.

Step 3: Convert moles of sodium chlorate to grams

Now that we know the number of moles of $NaClO_3$ required, we can convert this amount to grams using the molar mass of $NaClO_3$. The molar mass of $NaClO_3$ is given as 106.44 g/mol. We use the same formula as in step 1, but rearranged to solve for mass:

$Mass = Moles \times Molar\,Mass$

Plugging in the values, we get:

$Mass\,of\,NaClO_3 = 0.219\,mol \times 106.44\,g/mol = 23.31\,g$

Therefore, 23.31 g of $NaClO_3$ are needed to produce 10.5 g of $O_2$. This final conversion gives us the answer in the desired units (grams), which is a practical measure for laboratory work. By converting moles back to grams, we bridge the gap between the theoretical world of stoichiometry and the practical world of chemical experiments.

Conclusion

In summary, to calculate the mass of sodium chlorate ($NaClO_3$) required to form 10.5 g of oxygen ($O_2$), we followed these steps:

1. Converted the mass of oxygen to moles using its molar mass.
2. Used the stoichiometric ratio from the balanced chemical equation to find the moles of sodium chlorate.
3. Converted the moles of sodium chlorate to grams using its molar mass.

Through these calculations, we determined that 23.31 g of $NaClO_3$ is needed to produce 10.5 g of $O_2$. This exercise highlights the importance of stoichiometry in chemical calculations, allowing us to accurately predict the amounts of reactants and products involved in chemical reactions. Understanding these principles is fundamental to many areas of chemistry, from laboratory experiments to industrial processes.

The correct answer is not among the options provided (A. 34.9). The accurate calculation demonstrates the importance of careful stoichiometry and unit conversions in chemistry. Always double-check your work and ensure you are using the correct molar masses and stoichiometric ratios to arrive at the correct answer.

Stoichiometry is the cornerstone of quantitative chemistry, allowing us to predict the amounts of reactants and products involved in chemical reactions. Stoichiometric calculations are essential for chemists and students alike. In this comprehensive guide, we will delve into the principles of stoichiometry, focusing on how to calculate the mass of reactants needed to produce a specific amount of product, using the decomposition of sodium chlorate ($NaClO_3$) into oxygen ($O_2$) as our primary example. This process involves several critical steps, including balancing chemical equations, converting mass to moles, applying stoichiometric ratios, and converting moles back to mass. Mastering these steps is crucial for success in chemistry.

The balanced chemical equation for the decomposition of sodium chlorate is:

$2 NaClO_3(s) \rightarrow 2 NaCl(s) + 3 O_2(g)$

This equation tells us that 2 moles of solid sodium chlorate ($NaClO_3$) decompose to form 2 moles of solid sodium chloride ($NaCl$) and 3 moles of gaseous oxygen ($O_2$). The coefficients in this balanced equation are the key to understanding the molar relationships between the reactants and products. Without a balanced equation, stoichiometric calculations would be impossible, as the ratios would not accurately reflect the chemical reality of the reaction.

Understanding Molar Mass and Mole Conversions

The concept of molar mass is central to stoichiometric calculations. Molar mass, often denoted as M, is the mass of one mole of a substance, expressed in grams per mole (g/mol). The molar mass of a compound can be calculated by summing the atomic masses of all the atoms in the chemical formula. For example, the molar mass of oxygen ($O_2$) is approximately 32.00 g/mol (2 oxygen atoms x 16.00 g/mol each), and the molar mass of sodium chlorate ($NaClO_3$) is approximately 106.44 g/mol (22.99 g/mol for Na + 35.45 g/mol for Cl + 3 x 16.00 g/mol for O).

Mole conversions are the bridge between the macroscopic world (grams) and the microscopic world (moles). To convert the mass of a substance to moles, we use the formula:

$Moles = \frac{Mass}{Molar\,Mass}$

Conversely, to convert moles to mass, we use the formula:

$Mass = Moles \times Molar\,Mass$

These conversions are essential for stoichiometric calculations because balanced chemical equations describe the molar relationships between substances, not the mass relationships. Therefore, we must convert masses to moles, apply the stoichiometric ratios, and then convert moles back to mass to solve stoichiometric problems.

Applying Stoichiometric Ratios

The heart of stoichiometry lies in the use of stoichiometric ratios. Stoichiometric ratios are derived from the coefficients in the balanced chemical equation. These ratios provide the molar relationships between reactants and products. For example, in the decomposition of sodium chlorate:

$2 NaClO_3(s) \rightarrow 2 NaCl(s) + 3 O_2(g)$

The stoichiometric ratio between $NaClO_3$ and $O_2$ is 2:3. This means that for every 2 moles of $NaClO_3$ that decompose, 3 moles of $O_2$ are produced. We can use this ratio to calculate the amount of $NaClO_3$ needed to produce a specific amount of $O_2$, or vice versa.

To apply stoichiometric ratios, we set up a proportion using the balanced equation's coefficients. If we want to find out how many moles of $NaClO_3$ are needed to produce a certain number of moles of $O_2$, we use the following proportion:

$\frac{Moles\,of\,NaClO_3}{Moles\,of\,O_2} = \frac{2}{3}$

By plugging in the known number of moles of one substance, we can solve for the unknown number of moles of the other substance. This step is critical in stoichiometric calculations, as it allows us to move from the amount of one substance to the amount of another within a chemical reaction.

Step-by-Step Calculation: Mass of $NaClO_3$ Needed to Produce 10.5 g of $O_2$

Let's walk through a step-by-step calculation to determine the mass of sodium chlorate ($NaClO_3$) required to produce 10.5 g of oxygen ($O_2$).

Step 1: Convert the Mass of Oxygen to Moles

Given that we want to produce 10.5 g of $O_2$, and the molar mass of $O_2$ is 32.00 g/mol, we can convert the mass to moles using the formula:

$Moles = \frac{Mass}{Molar\,Mass}$

$Moles\,of\,O_2 = \frac{10.5\,g}{32.00\,g/mol} = 0.328\,mol$

Thus, 10.5 g of $O_2$ is equivalent to 0.328 moles. This conversion is the crucial first step, as it allows us to work with the molar ratios from the balanced chemical equation.

Step 2: Use the Stoichiometric Ratio to Find Moles of Sodium Chlorate

From the balanced equation:

$2 NaClO_3(s) \rightarrow 2 NaCl(s) + 3 O_2(g)$

we know that the stoichiometric ratio between $NaClO_3$ and $O_2$ is 2:3. Therefore, to find the moles of $NaClO_3$ needed, we set up the proportion:

$\frac{Moles\,of\,NaClO_3}{Moles\,of\,O_2} = \frac{2}{3}$

Plugging in the moles of $O_2$ we calculated in Step 1, we get:

$\frac{Moles\,of\,NaClO_3}{0.328\,mol} = \frac{2}{3}$

Solving for moles of $NaClO_3$:

$Moles\,of\,NaClO_3 = \frac{2}{3} \times 0.328\,mol = 0.219\,mol$

So, 0.219 moles of $NaClO_3$ are required to produce 0.328 moles of $O_2$. This step demonstrates the power of stoichiometric ratios in relating the amounts of different substances in a chemical reaction.

Step 3: Convert Moles of Sodium Chlorate to Grams

Finally, we convert the moles of $NaClO_3$ to grams using the molar mass of $NaClO_3$, which is 106.44 g/mol:

$Mass = Moles \times Molar\,Mass$

$Mass\,of\,NaClO_3 = 0.219\,mol \times 106.44\,g/mol = 23.31\,g$

Therefore, 23.31 g of $NaClO_3$ are needed to produce 10.5 g of $O_2$. This final conversion gives us the answer in grams, which is a practical unit for measuring chemicals in the laboratory.

Common Mistakes in Stoichiometric Calculations

While stoichiometric calculations are systematic, they are also prone to errors if not performed carefully. Some common mistakes include:

1. Not Balancing the Chemical Equation: An unbalanced equation will lead to incorrect stoichiometric ratios and, consequently, incorrect results.
2. Using Incorrect Molar Masses: Using the wrong molar mass for a substance will throw off the mole conversions and subsequent calculations. Always double-check the molar masses you use.
3. Misapplying Stoichiometric Ratios: Applying the stoichiometric ratios incorrectly, such as inverting the ratio or using the wrong coefficients, will lead to errors.
4. Incorrect Unit Conversions: Failing to convert between grams and moles correctly is a frequent source of errors. Always ensure you are using the correct conversion factors.
5. Rounding Errors: Rounding intermediate values prematurely can lead to significant errors in the final result. It's best to carry extra digits through the calculation and round only at the end.

By being aware of these common pitfalls, you can improve your accuracy in stoichiometric calculations and avoid costly mistakes.

Advanced Stoichiometry: Limiting Reactants and Percent Yield

While the previous example focused on a straightforward stoichiometric calculation, many real-world chemical reactions involve more complex scenarios, such as limiting reactants and percent yield. Understanding these concepts is crucial for advanced stoichiometric problem-solving.

Limiting Reactants

In a chemical reaction, the limiting reactant is the reactant that is completely consumed, thus determining the maximum amount of product that can be formed. The other reactants are said to be in excess because there is more of them than is needed to react with the limiting reactant. Identifying the limiting reactant is crucial for accurately calculating the amount of product formed in a reaction.

To determine the limiting reactant, you must:

1. Calculate the moles of each reactant.
2. Use the stoichiometric ratios to determine how much of each reactant is needed to react completely with the other reactants.
3. Identify which reactant is present in the smallest amount relative to its stoichiometric requirement. This is the limiting reactant.

Percent Yield

The theoretical yield is the maximum amount of product that can be formed in a reaction, calculated based on the stoichiometry and the amount of limiting reactant. However, in practice, the actual yield (the amount of product actually obtained) is often less than the theoretical yield due to factors such as incomplete reactions, side reactions, and loss of product during purification.

The percent yield is a measure of the efficiency of a reaction, calculated as:

$Percent\,Yield = (\frac{Actual\,Yield}{Theoretical\,Yield}) \times 100\%$

Understanding percent yield is essential for optimizing chemical reactions and assessing the success of a synthesis.

Stoichiometry in Real-World Applications

Stoichiometry is not just a theoretical concept; it has numerous practical applications in various fields, including:

• Industrial Chemistry: Stoichiometry is used to optimize chemical processes, maximize product yield, and minimize waste in the chemical industry.
• Pharmaceuticals: Stoichiometric calculations are crucial for synthesizing drugs and ensuring the correct dosage in medications.
• Environmental Science: Stoichiometry is used to analyze pollution, calculate the amounts of pollutants in the environment, and design remediation strategies.
• Analytical Chemistry: Stoichiometry is fundamental to quantitative analysis, where the amounts of substances in a sample are determined.
• Cooking: Surprisingly, stoichiometry even plays a role in cooking, where recipes involve specific ratios of ingredients to achieve the desired result.

Conclusion: The Power of Stoichiometric Thinking

Stoichiometry is a fundamental tool in chemistry that allows us to quantify the relationships between reactants and products in chemical reactions. By mastering the principles of stoichiometry, including balancing equations, mole conversions, stoichiometric ratios, limiting reactants, and percent yield, you gain the ability to predict and control chemical reactions. Whether you are a student learning the basics or a professional chemist working on cutting-edge research, stoichiometric thinking is essential for success in the chemical sciences. This article has provided a comprehensive guide to stoichiometric calculations, focusing on the decomposition of sodium chlorate to produce oxygen as a prime example. By understanding and applying these principles, you can confidently tackle a wide range of chemical problems and appreciate the power and elegance of stoichiometry.

Sodium chlorate ($NaClO_3$) decomposition is a classic example used to illustrate the principles of stoichiometry. Sodium chlorate is a strong oxidizer, and its decomposition reaction produces oxygen gas, which has numerous applications in industry, medicine, and research. The reaction is typically carried out by heating sodium chlorate, often in the presence of a catalyst, to lower the activation energy and increase the reaction rate. This process provides an excellent context for understanding and applying stoichiometric calculations. In this article, we will explore the decomposition of sodium chlorate in detail, focusing on the stoichiometric relationships, reaction conditions, and practical applications.

The balanced chemical equation for the decomposition of sodium chlorate is:

$2 NaClO_3(s) \rightarrow 2 NaCl(s) + 3 O_2(g)$

This equation is the foundation for all stoichiometric calculations involving this reaction. It tells us that two moles of solid sodium chlorate decompose to form two moles of solid sodium chloride and three moles of gaseous oxygen. The coefficients in this equation are the molar ratios that we use to relate the amounts of reactants and products. Understanding this balanced equation is crucial for any quantitative analysis of the reaction.

Understanding the Reaction Mechanism

The decomposition of sodium chlorate is a thermal decomposition reaction, meaning it is driven by heat. When heated, sodium chlorate breaks down into sodium chloride and oxygen gas. The reaction mechanism is complex and can be influenced by several factors, including temperature, pressure, and the presence of catalysts. Catalysts, such as manganese dioxide ($MnO_2$), can significantly lower the activation energy of the reaction, allowing it to proceed at a lower temperature and a faster rate.

The reaction mechanism can be thought of as a series of elementary steps, although the exact details are not always fully understood. At a high level, the process involves the breaking of chemical bonds within the sodium chlorate molecule and the formation of new bonds to create sodium chloride and oxygen gas. The energy required to break the initial bonds is supplied by the heat, and the overall reaction is exothermic, meaning it releases energy in the form of heat.

Factors Affecting the Reaction Rate

Several factors can influence the rate of the sodium chlorate decomposition reaction:

1. Temperature: Higher temperatures generally lead to faster reaction rates because the molecules have more kinetic energy, making it easier to overcome the activation energy barrier.
2. Catalysts: Catalysts lower the activation energy of the reaction, allowing it to proceed faster at a given temperature. Manganese dioxide ($MnO_2$) is a common catalyst used in this reaction.
3. Surface Area: If the sodium chlorate is finely divided (i.e., has a high surface area), the reaction can occur more rapidly because there is more contact between the reactant and the heat source.
4. Pressure: Pressure can also affect the reaction rate, particularly for gaseous products like oxygen. Higher pressures can increase the concentration of reactants and products, potentially affecting the reaction rate.

Understanding these factors is essential for optimizing the reaction conditions in both laboratory and industrial settings. By carefully controlling these variables, chemists can ensure efficient and safe oxygen production.

Stoichiometric Calculations: A Practical Example

Let's consider a practical example of stoichiometric calculations involving the decomposition of sodium chlorate. Suppose we want to produce 50.0 grams of oxygen gas ($O_2$). How much sodium chlorate ($NaClO_3$) do we need to start with?

Step 1: Convert the Mass of Oxygen to Moles

We start by converting the mass of oxygen to moles using the molar mass of $O_2$, which is approximately 32.00 g/mol:

$Moles\,of\,O_2 = \frac{Mass}{Molar\,Mass} = \frac{50.0\,g}{32.00\,g/mol} = 1.56\,mol$

So, 50.0 grams of $O_2$ is equivalent to 1.56 moles.

Step 2: Use the Stoichiometric Ratio to Find Moles of Sodium Chlorate

From the balanced equation:

$2 NaClO_3(s) \rightarrow 2 NaCl(s) + 3 O_2(g)$

we see that the stoichiometric ratio between $NaClO_3$ and $O_2$ is 2:3. We can set up a proportion to find the moles of $NaClO_3$ needed:

$\frac{Moles\,of\,NaClO_3}{Moles\,of\,O_2} = \frac{2}{3}$

Plugging in the moles of $O_2$ we calculated:

$\frac{Moles\,of\,NaClO_3}{1.56\,mol} = \frac{2}{3}$

Solving for moles of $NaClO_3$:

$Moles\,of\,NaClO_3 = \frac{2}{3} \times 1.56\,mol = 1.04\,mol$

Therefore, we need 1.04 moles of $NaClO_3$ to produce 1.56 moles of $O_2$.

Step 3: Convert Moles of Sodium Chlorate to Grams

Finally, we convert the moles of $NaClO_3$ to grams using the molar mass of $NaClO_3$, which is approximately 106.44 g/mol:

$Mass\,of\,NaClO_3 = Moles \times Molar\,Mass = 1.04\,mol \times 106.44\,g/mol = 110.70\,g$

Thus, we need approximately 110.70 grams of sodium chlorate to produce 50.0 grams of oxygen gas. This step-by-step calculation illustrates the practical application of stoichiometry in determining the amounts of reactants needed for a specific chemical reaction.

Safety Considerations

Working with sodium chlorate and its decomposition reaction requires careful attention to safety. Sodium chlorate is a strong oxidizer and can react violently with combustible materials, leading to fire or explosion. Therefore, it should be handled with caution and stored away from flammable substances. When performing the decomposition reaction, it is crucial to control the temperature and use appropriate safety equipment, such as gloves, goggles, and a fume hood to vent the oxygen gas.

Applications of Oxygen Production from Sodium Chlorate

The oxygen produced from the decomposition of sodium chlorate has various applications, including:

1. Laboratory Experiments: Oxygen gas is often produced in the laboratory for various experiments, such as combustion reactions and gas analysis.
2. Medical Use: Oxygen is used in medical settings for patients with respiratory problems.
3. Industrial Processes: Oxygen is used in many industrial processes, such as welding, metal cutting, and the production of steel.
4. Emergency Oxygen Supply: Small-scale oxygen generators based on sodium chlorate decomposition can be used in emergency situations, such as in airplanes or submarines.

The versatility of oxygen makes the decomposition of sodium chlorate a valuable method for its production in various settings.

Conclusion: Sodium Chlorate Decomposition as a Stoichiometry Case Study

The decomposition of sodium chlorate is a classic example of a chemical reaction that perfectly illustrates the principles of stoichiometry. By understanding the balanced chemical equation, reaction mechanism, and factors affecting the reaction rate, we can perform accurate stoichiometric calculations to determine the amounts of reactants and products involved. This reaction has practical applications in various fields, from laboratory experiments to industrial processes. However, it is essential to handle sodium chlorate with caution due to its oxidizing properties. Through this deep dive into sodium chlorate decomposition, we have reinforced the importance of stoichiometry in chemistry and its real-world relevance.