Number of Moles of NiS2 and O2

Oct 13, 2025 by vinooalluri.com 80 views

Understanding the Reaction between NiS2 and O2: A Comprehensive Analysis

In this article, we will delve into the reaction between nickel(II) sulfide ($$NiS_2$$) and oxygen ($$O_2$$), which results in the formation of nickel(II) oxide ($$NiO$$) and sulfur dioxide ($$SO_2$$). The balanced chemical equation for this reaction is:

2 NiS_2(s) + 5 O_2(g) \longrightarrow 2 NiO(s) + 4 SO_2(g)

To fully comprehend this reaction, we need to determine the limiting reactant, calculate the theoretical yield of $$NiO$$, and finally, assess the percent yield of the reaction.

To identify the limiting reactant, we need to calculate the number of moles of each reactant required to produce the desired product. Let's assume we have 100 grams of $$NiS_2$$ and 100 grams of $$O_2$$.

Molar Mass of NiS2 and O2

The molar mass of $$NiS_2$$ is calculated as follows:

NiS_2 = 58.69 (Ni) + 2 \times 32.07 (S) = 122.83 g/mol

The molar mass of $$O_2$$ is calculated as follows:

O_2 = 2 \times 16.00 (O) = 32.00 g/mol

Number of Moles of NiS2 and O2

Now, let's calculate the number of moles of each reactant:

moles (NiS_2) = \frac{100 g}{122.83 g/mol} = 0.814 mol

moles (O_2) = \frac{100 g}{32.00 g/mol} = 3.125 mol

Moles of O2 Required

According to the balanced chemical equation, 5 moles of $$O_2$$ are required to produce 2 moles of $$NiO$$. Therefore, the number of moles of $$O_2$$ required is:

moles (O_2)_{required} = \frac{2 mol (NiO)}{5 mol (O_2)} \times 0.814 mol (NiS_2) = 0.326 mol

Since we have 3.125 moles of $$O_2$$, which is more than the required amount, $$O_2$$ is in excess.

Moles of NiS2 Required

According to the balanced chemical equation, 2 moles of $$NiS_2$$ are required to produce 2 moles of $$NiO$$. Therefore, the number of moles of $$NiS_2$$ required is:

moles (NiS_2)_{required} = \frac{2 mol (NiO)}{2 mol (NiS_2)} \times 0.814 mol (NiS_2) = 0.814 mol

Since we have 0.814 moles of $$NiS_2$$, which is equal to the required amount, $$NiS_2$$ is the limiting reactant.

Now that we have identified the limiting reactant, we can calculate the theoretical yield of $$NiO$$.

The molar mass of $$NiO$$ is calculated as follows:

NiO = 58.69 (Ni) + 16.00 (O) = 74.69 g/mol

The number of moles of $$NiO$$ produced is:

moles (NiO) = \frac{2 mol (NiO)}{2 mol (NiS_2)} \times 0.814 mol (NiS_2) = 0.814 mol

The mass of $$NiO$$ produced is:

mass (NiO) = moles (NiO) \times molar mass (NiO) = 0.814 mol \times 74.69 g/mol = 60.93 g

To calculate the percent yield of $$NiO$$, we need to know the actual yield of $$NiO$$.

Let's assume the actual yield of $$NiO$$ is 50 grams.

The percent yield of $$NiO$$ is calculated as follows:

percent yield = \frac{actual yield}{theoretical yield} \times 100\% = \frac{50 g}{60.93 g} \times 100\% = 81.8\%

In conclusion, the limiting reactant in the reaction between $$NiS_2$$ and $$O_2$$ is $$NiS_2$$. The theoretical yield of $$NiO$$ is 60.93 grams, and the percent yield of $$NiO$$ is 81.8%. This analysis provides a comprehensive understanding of the reaction and its limitations.

Chemistry: An Atoms First Approach, 2nd Edition, by Steven S. Zumdahl
General Chemistry: Principles and Modern Applications, 10th Edition, by Linus Pauling
Q&A: Understanding the Reaction between NiS2 and O2

In our previous article, we delved into the reaction between nickel(II) sulfide ($$NiS_2$$) and oxygen ($$O_2$$), which results in the formation of nickel(II) oxide ($$NiO$$) and sulfur dioxide ($$SO_2$$). In this article, we will answer some frequently asked questions about this reaction to provide a deeper understanding of the process.

Q: What is the purpose of determining the limiting reactant?

A: Determining the limiting reactant is crucial in understanding the reaction's outcome. The limiting reactant is the reactant that is consumed first, and it determines the maximum amount of product that can be formed. In the case of the reaction between $$NiS_2$$ and $$O_2$$, $$NiS_2$$ is the limiting reactant.

Q: How do you calculate the number of moles of each reactant?

A: To calculate the number of moles of each reactant, you need to know the molar mass of each reactant. The molar mass is the sum of the atomic masses of the elements in the compound. For example, the molar mass of $$NiS_2$$ is 122.83 g/mol, and the molar mass of $$O_2$$ is 32.00 g/mol.

Q: What is the difference between theoretical yield and actual yield?

A: Theoretical yield is the maximum amount of product that can be formed based on the limiting reactant, while actual yield is the amount of product that is actually obtained. In the case of the reaction between $$NiS_2$$ and $$O_2$$, the theoretical yield of $$NiO$$ is 60.93 grams, while the actual yield is 50 grams.

Q: How do you calculate the percent yield of a product?

A: To calculate the percent yield of a product, you need to know the actual yield and the theoretical yield. The percent yield is calculated as follows:

percent yield = \frac{actual yield}{theoretical yield} \times 100\%

Q: What factors can affect the percent yield of a product?

A: Several factors can affect the percent yield of a product, including:

Impurities: Impurities in the reactants can affect the reaction's outcome and reduce the percent yield.
Inaccurate measurements: Inaccurate measurements of the reactants can affect the reaction's outcome and reduce the percent yield.
Inadequate reaction conditions: Inadequate reaction conditions, such as temperature or pressure, can affect the reaction's outcome and reduce the percent yield.

Q: How can you improve the percent yield of a product?

A: To improve the percent yield of a product, you can:

Use high-purity reactants: Using high-purity reactants can reduce the effect of impurities on the reaction's outcome.
Accurately measure the reactants: Accurately measuring the reactants can ensure that the reaction is carried out under the correct conditions.
Optimize the reaction conditions: Optimizing the reaction conditions, such as temperature or pressure, can improve the reaction's outcome and increase the percent yield.

In conclusion, understanding the reaction between $$NiS_2$$ and $$O_2$$ requires a comprehensive analysis of the reaction's outcome, including the limiting reactant, theoretical yield, and percent yield. By answering frequently asked questions about this reaction, we can gain a deeper understanding of the process and improve the percent yield of the product.

Chemistry: An Atoms First Approach, 2nd Edition, by Steven S. Zumdahl
General Chemistry: Principles and Modern Applications, 10th Edition, by Linus Pauling