Limiting Reactant Calculation For Nickel(II) Oxide Synthesis

Jul 11, 2025 by qnaunigon 61 views

Understanding Limiting Reactants in Nickel(II) Oxide Synthesis

Introduction

In the realm of chemistry, stoichiometry plays a pivotal role in understanding the quantitative relationships between reactants and products in chemical reactions. One crucial concept within stoichiometry is the limiting reactant, which dictates the maximum amount of product that can be formed in a given reaction. This article delves into a scenario involving the synthesis of nickel(II) oxide (NiO) from nickel (Ni) and oxygen ( $O_2$ ), focusing on identifying the limiting reactant and determining the resulting quantities of reactants and products. By examining the balanced chemical equation and the initial amounts of reactants, we can unravel the intricacies of this chemical process and gain a deeper understanding of stoichiometric principles. Let’s embark on this chemical journey to explore the fascinating world of limiting reactants and their impact on chemical reactions. We'll dissect the reaction, identify the limiting reactant, and calculate the amounts of reactants and products present after the reaction reaches completion.

Problem Statement: Nickel(II) Oxide Synthesis

A scientist initiates a reaction with 7.81 moles of nickel (Ni) and 4.51 moles of oxygen ( $O_2$ ) to synthesize nickel(II) oxide (NiO). The balanced chemical equation for this reaction is:

$2 Ni + O_2 \rightarrow 2 NiO$

The primary objective is to determine the maximum amount of nickel(II) oxide (NiO) that can be formed and the amounts of each substance present after the reaction is completed. This problem necessitates a thorough understanding of stoichiometry, specifically the concept of the limiting reactant. The limiting reactant is the reactant that is completely consumed in a chemical reaction, thus dictating the maximum amount of product that can be formed. To solve this, we'll first identify the limiting reactant by comparing the mole ratios of the reactants to the stoichiometric coefficients in the balanced equation. Then, we'll use the limiting reactant to calculate the moles of product formed and the moles of excess reactant remaining. This step-by-step approach will provide a clear and concise solution to the problem, highlighting the importance of stoichiometry in predicting the outcomes of chemical reactions.

Step 1: Identifying the Limiting Reactant

The limiting reactant is the key to determining the maximum yield of a product in a chemical reaction. To identify the limiting reactant, we need to compare the mole ratios of the reactants to the stoichiometric coefficients in the balanced chemical equation. The balanced equation for the synthesis of nickel(II) oxide is:

$2 Ni + O_2 \rightarrow 2 NiO$

This equation tells us that 2 moles of nickel (Ni) react with 1 mole of oxygen ( $O_2$ ) to produce 2 moles of nickel(II) oxide (NiO). We are given that we have 7.81 moles of Ni and 4.51 moles of $O_2$ . To determine which reactant is limiting, we can calculate how many moles of one reactant are required to react completely with the given amount of the other reactant.

Let's start by calculating how many moles of $O_2$ are needed to react with 7.81 moles of Ni. According to the balanced equation, the mole ratio of Ni to $O_2$ is 2:1. Therefore, the moles of $O_2$ required can be calculated as follows:

Moles of $O_2$ required = (7.81 moles Ni) * (1 mole $O_2$ / 2 moles Ni) = 3.905 moles $O_2$

We have 4.51 moles of $O_2$ , which is more than the 3.905 moles required to react with all the Ni. This indicates that Ni is the limiting reactant because it will be completely consumed before all the $O_2$ is used up. Alternatively, we can calculate how many moles of Ni are required to react with 4.51 moles of $O_2$ :

Moles of Ni required = (4.51 moles $O_2$ ) * (2 moles Ni / 1 mole $O_2$ ) = 9.02 moles Ni

Since we only have 7.81 moles of Ni, which is less than the 9.02 moles required to react with all the $O_2$ , Ni is confirmed as the limiting reactant. Identifying the limiting reactant is a crucial step in stoichiometry as it dictates the maximum amount of product that can be formed. In this case, the amount of Ni present will determine the amount of NiO produced. Understanding the mole ratios and performing these calculations are fundamental skills in chemistry, enabling us to predict the outcomes of chemical reactions accurately.

Step 2: Calculating Moles of NiO Produced

Now that we've identified nickel (Ni) as the limiting reactant, we can calculate the maximum amount of nickel(II) oxide (NiO) that can be produced. The balanced chemical equation provides the stoichiometric relationship between Ni and NiO:

$2 Ni + O_2 \rightarrow 2 NiO$

This equation shows that 2 moles of Ni react to produce 2 moles of NiO. Therefore, the mole ratio of Ni to NiO is 1:1. This simple ratio makes the calculation straightforward. We start with 7.81 moles of Ni, the limiting reactant. Using the mole ratio, we can determine the moles of NiO produced:

Moles of NiO produced = (7.81 moles Ni) * (2 moles NiO / 2 moles Ni) = 7.81 moles NiO

Thus, the maximum amount of NiO that can be produced from 7.81 moles of Ni is 7.81 moles. This calculation highlights the direct relationship between the limiting reactant and the product formed. The limiting reactant is completely consumed, and its initial amount dictates the theoretical yield of the product. In this case, the 7.81 moles of Ni will yield 7.81 moles of NiO. Understanding these stoichiometric relationships is crucial in chemistry for predicting reaction outcomes and optimizing chemical processes. The concept of the limiting reactant is not only important in theoretical calculations but also in practical applications, such as industrial chemical synthesis, where maximizing product yield is essential for economic efficiency. This calculation demonstrates the power of stoichiometry in providing quantitative insights into chemical reactions.

Step 3: Calculating Moles of Excess Reactant Remaining

After determining the amount of nickel(II) oxide (NiO) produced, the next step is to calculate the amount of the excess reactant remaining after the reaction is complete. In this case, oxygen ( $O_2$ ) is the excess reactant because we have more $O_2$ than is needed to react with all the nickel (Ni). To calculate the remaining amount of $O_2$ , we first need to determine how much $O_2$ reacted with the Ni. From the balanced chemical equation:

$2 Ni + O_2 \rightarrow 2 NiO$

We know that 2 moles of Ni react with 1 mole of $O_2$ . Since Ni is the limiting reactant, all 7.81 moles of Ni will react. We can use the stoichiometric ratio to find out how many moles of $O_2$ reacted:

Moles of $O_2$ reacted = (7.81 moles Ni) * (1 mole $O_2$ / 2 moles Ni) = 3.905 moles $O_2$

So, 3.905 moles of $O_2$ reacted with the 7.81 moles of Ni. We started with 4.51 moles of $O_2$ . To find the moles of $O_2$ remaining, we subtract the amount reacted from the initial amount:

Moles of $O_2$ remaining = Initial moles of $O_2$ - Moles of $O_2$ reacted Moles of $O_2$ remaining = 4.51 moles - 3.905 moles = 0.605 moles $O_2$

Therefore, 0.605 moles of $O_2$ remain after the reaction is complete. This calculation is essential for a complete understanding of the reaction outcome. It demonstrates that not all reactants are necessarily consumed in a chemical reaction; the limiting reactant dictates the extent of the reaction, and excess reactants remain. This concept is particularly important in industrial chemistry, where optimizing the use of reactants can significantly impact the efficiency and cost-effectiveness of a process. Understanding how to calculate the amount of excess reactant remaining provides valuable insights into reaction stoichiometry and helps in predicting the final composition of the reaction mixture.

Final Answer: Moles of Each Substance Present

Having gone through the step-by-step calculations, we can now summarize the final amounts of each substance present after the reaction between nickel (Ni) and oxygen ( $O_2$ ) to form nickel(II) oxide (NiO) has reached completion. We started with 7.81 moles of Ni and 4.51 moles of $O_2$ . We identified Ni as the limiting reactant, which means it is completely consumed in the reaction. Therefore, the moles of Ni remaining are:

Moles of Ni remaining = 0 moles

Next, we calculated the amount of nickel(II) oxide (NiO) produced. Since the mole ratio of Ni to NiO is 1:1, and we started with 7.81 moles of Ni, the moles of NiO produced are:

Moles of NiO produced = 7.81 moles

Finally, we determined the amount of oxygen ( $O_2$ ) remaining. We calculated that 3.905 moles of $O_2$ reacted with the Ni, and we started with 4.51 moles of $O_2$ . Thus, the moles of $O_2$ remaining are:

Moles of $O_2$ remaining = 0.605 moles

In summary, after the reaction is complete, we have:

0 moles of Ni
7.81 moles of NiO
0.605 moles of $O_2$

This comprehensive analysis provides a clear picture of the reaction outcome, highlighting the crucial role of the limiting reactant in determining the final composition of the reaction mixture. Understanding these stoichiometric relationships is fundamental to mastering chemistry and its applications in various fields, from chemical synthesis to environmental science. The ability to accurately predict the amounts of reactants and products in a chemical reaction is a cornerstone of chemical knowledge and a valuable skill for any aspiring scientist or engineer.

Conclusion

In conclusion, this exercise has demonstrated the importance of understanding stoichiometry and the concept of the limiting reactant in predicting the outcome of chemical reactions. By carefully analyzing the balanced chemical equation and the initial amounts of reactants, we were able to determine that nickel (Ni) was the limiting reactant in the synthesis of nickel(II) oxide (NiO). This allowed us to calculate the maximum amount of NiO produced (7.81 moles) and the amount of excess oxygen ( $O_2$ ) remaining (0.605 moles) after the reaction was complete. The limiting reactant concept is not just a theoretical exercise; it has significant practical implications in various fields, including industrial chemistry, where optimizing reaction yields is crucial for economic efficiency. Understanding how to identify the limiting reactant and calculate the amounts of products and excess reactants is a fundamental skill for chemists and chemical engineers. This problem illustrates the power of stoichiometry in providing quantitative insights into chemical reactions and its essential role in chemical problem-solving. Mastering these concepts provides a solid foundation for further studies in chemistry and related disciplines. The ability to accurately predict reaction outcomes is a key aspect of chemical knowledge and a valuable tool for addressing real-world challenges in various fields.