Stoichiometry Of Sucrose Combustion Calculating Moles And Limiting Reactant

#Stoichiometry plays a crucial role# in understanding chemical reactions, particularly combustion. This article delves into the stoichiometry of sucrose ($C_{12}H_{22}O_{11}$) combustion, a common process that releases energy. We will analyze the balanced chemical equation, calculate the moles of reactants, identify the limiting reactant, and determine the theoretical yield of products. This comprehensive exploration will provide a solid understanding of the quantitative aspects of this reaction. This exploration is crucial for students, chemists, and anyone interested in understanding the chemical reactions that power our world.

Balanced Chemical Equation for Sucrose Combustion

The balanced chemical equation is the foundation of stoichiometric calculations. It provides the molar ratios between reactants and products, which are essential for determining the quantities of substances involved in a reaction. The balanced equation for the combustion of sucrose is:

$C_{12}H_{22}O_{11} + 12O_2 ightarrow 12CO_2 + 11H_2O$

This equation tells us that one mole of sucrose reacts with twelve moles of oxygen to produce twelve moles of carbon dioxide and eleven moles of water. This molar ratio is the key to solving stoichiometry problems. Understanding the stoichiometric coefficients is crucial for accurately predicting the outcome of the reaction. The coefficients not only represent the number of moles but also provide a ratio of the number of molecules involved in the reaction. A balanced chemical equation adheres to the law of conservation of mass, ensuring that the number of atoms of each element is the same on both sides of the equation.

This balanced equation allows us to quantitatively analyze the reaction, predicting the amounts of reactants needed and the amounts of products formed. For instance, if we know the amount of sucrose, we can use the stoichiometric ratio to determine the amount of oxygen required for complete combustion. Conversely, if we know the amount of oxygen available, we can calculate the maximum amount of sucrose that can react. The coefficients in the balanced equation act as conversion factors, allowing us to move between moles of different substances within the reaction. This foundational understanding is essential for various applications, from industrial chemical processes to understanding biological energy production. The balanced equation also provides insights into the mechanism of the reaction, although it does not explicitly show the reaction pathway.

Calculating Moles of Sucrose and Oxygen

To determine how many moles of sucrose are available for the reaction, we need to convert the given mass of sucrose (10.0 g) to moles using its molar mass. The molar mass of sucrose ($C_{12}H_{22}O_{11}$) is calculated as:

(12 × 12.01 g/mol) + (22 × 1.01 g/mol) + (11 × 16.00 g/mol) = 342.30 g/mol

Now, we can calculate the moles of sucrose:

Moles of sucrose = Mass of sucrose / Molar mass of sucrose Moles of sucrose = 10.0 g / 342.30 g/mol = 0.0292 mol

Similarly, we need to calculate the moles of oxygen ($O_2$) using its molar mass, which is 32.00 g/mol. The given mass of oxygen is 8.0 g.

Moles of oxygen = Mass of oxygen / Molar mass of oxygen Moles of oxygen = 8.0 g / 32.00 g/mol = 0.25 mol

These calculations are fundamental in stoichiometry, as they allow us to work with the mole ratios provided by the balanced equation. The mole is a central unit in chemistry, representing a specific number of molecules (Avogadro's number, 6.022 × 10^23). By converting grams to moles, we can directly relate the amounts of different substances in a chemical reaction. These calculations form the basis for determining the limiting reactant and the theoretical yield of the reaction. The accuracy of these calculations is paramount, as any errors will propagate through subsequent calculations. Understanding molar mass and its application in converting between mass and moles is a crucial skill in chemistry.

Determining the Limiting Reactant

The limiting reactant is the reactant that is completely consumed in a chemical reaction, thereby determining the maximum amount of product that can be formed. To identify the limiting reactant, we need to compare the mole ratio of the reactants to the stoichiometric ratio from the balanced equation.

From the balanced equation, the stoichiometric ratio of sucrose to oxygen is 1:12. This means that one mole of sucrose requires twelve moles of oxygen for complete combustion. Now, let's compare the available moles of sucrose and oxygen to this ratio.

We have 0.0292 mol of sucrose and 0.25 mol of oxygen. To determine the limiting reactant, we can calculate how much oxygen is needed to react completely with 0.0292 mol of sucrose:

Moles of oxygen needed = Moles of sucrose × Stoichiometric ratio (O2 to sucrose) Moles of oxygen needed = 0.0292 mol × 12 = 0.3504 mol

Since we only have 0.25 mol of oxygen, which is less than the 0.3504 mol required, oxygen is the limiting reactant. Alternatively, we can calculate how much sucrose is needed to react completely with 0.25 mol of oxygen:

Moles of sucrose needed = Moles of oxygen / Stoichiometric ratio (O2 to sucrose) Moles of sucrose needed = 0.25 mol / 12 = 0.0208 mol

Since we have 0.0292 mol of sucrose, which is more than the 0.0208 mol required, oxygen is again identified as the limiting reactant. The limiting reactant concept is critical in optimizing chemical reactions. By identifying the limiting reactant, we can ensure that the more expensive or valuable reactant is fully utilized, maximizing product yield and minimizing waste. In industrial processes, careful control of reactant ratios is essential for economic efficiency. The limiting reactant not only affects the amount of product formed but also influences the reaction rate and the overall reaction kinetics.

Calculating Theoretical Yield

The theoretical yield is the maximum amount of product that can be formed from a given amount of limiting reactant, assuming the reaction goes to completion with no losses. In this case, oxygen is the limiting reactant, so we will use its moles (0.25 mol) to calculate the theoretical yield of carbon dioxide ($CO_2$).

From the balanced equation, 12 moles of $CO_2$ are produced for every 12 moles of $O_2$ reacted. This simplifies the calculation since the mole ratio is 1:1.

Moles of $CO_2$ produced = Moles of $O_2$ reacted × Stoichiometric ratio ($CO_2$ to $O_2$) Moles of $CO_2$ produced = 0.25 mol × (12 mol $CO_2$ / 12 mol $O_2$) = 0.25 mol

Now, we need to convert moles of $CO_2$ to grams using its molar mass. The molar mass of $CO_2$ is:

(1 × 12.01 g/mol) + (2 × 16.00 g/mol) = 44.01 g/mol

Theoretical yield of $CO_2$ = Moles of $CO_2$ × Molar mass of $CO_2$ Theoretical yield of $CO_2$ = 0.25 mol × 44.01 g/mol = 11.0025 g

Therefore, the theoretical yield of carbon dioxide is approximately 11.00 g. The theoretical yield represents an ideal scenario, assuming no side reactions or losses during the process. In reality, the actual yield is often less than the theoretical yield due to factors such as incomplete reactions, loss of product during purification, and side reactions. The percentage yield, which is the ratio of the actual yield to the theoretical yield, is a common measure of the efficiency of a chemical reaction. Calculating the theoretical yield is a critical step in planning and optimizing chemical reactions, providing a benchmark for the expected amount of product.

Conclusion

In summary, understanding the stoichiometry of sucrose combustion involves balancing the chemical equation, calculating the moles of reactants, identifying the limiting reactant, and determining the theoretical yield. In this specific scenario, with 10.0 g of sucrose and 8.0 g of oxygen, oxygen was identified as the limiting reactant, and the theoretical yield of carbon dioxide was calculated to be approximately 11.00 g. This comprehensive analysis highlights the importance of stoichiometry in quantitative chemistry, providing the tools to predict and optimize chemical reactions. Mastering these concepts is crucial for success in chemistry and related fields. Stoichiometry not only provides a quantitative framework for understanding chemical reactions but also serves as a foundation for more advanced topics such as chemical kinetics and thermodynamics. The principles discussed here are applicable to a wide range of chemical reactions, making this analysis a valuable learning experience. By understanding the stoichiometry of sucrose combustion, we gain insights into the fundamental processes that underpin energy production and chemical transformations in various contexts.