Problem 86 - Chemistry: An Atoms-Focused Approach

Question

Black Powder Gunpowder, known also as black powder, is generally thought to have been invented by the Chinese in the 9 th century. The first description in English was given by Roger Bacon in the 13 th century. The basic formulation has not changed since then: \(40 \%\) potassium nitrate, \(30 \%\) carbon, and \(30 \%\) sulfur by weight. The products of the reaction that provide the explosive force when the powder is ignited are three gases: carbon monoxide, carbon dioxide, and nitrogen. An additional product is potassium sulfite. Consider a sample of \(100 \mathrm{g}\) of black powder. a. Which reagent is limiting? b. How much of each of the reagents in excess is left over after the explosion takes place?

Step-by-Step Solution

Answer

Answer: The limiting reagent is potassium nitrate (KNO₃). After the explosion, there will be 15.8g of carbon and 17.3g of sulfur left over.

Step 1: 1. Write the balanced chemical equation for the black powder reaction

First, we need to identify the components of black powder: potassium nitrate (KNO₃), carbon (C), and sulfur (S). The products of the reaction are carbon monoxide (CO), carbon dioxide (CO₂), nitrogen (N₂), and potassium sulfite (K₂SO₃). The balanced chemical equation for the black powder reaction is: \(2 \mathrm{KNO}_{3} + 3 \mathrm{C} + \mathrm{S} \rightarrow 3 \mathrm{CO} + \mathrm{CO}_{2} + \mathrm{K}_{2}\mathrm{SO}_{3} + \frac{1}{2}\mathrm{N}_{2}\)

Step 2: 2. Determine the mass of each reactant

We are given a \(100\mathrm{g}\) sample of black powder. We need to find out the mass of each of the components: KNO₃, C, and S. - KNO₃: \(40 \% \Rightarrow 40\mathrm{g}\) - C: \(30 \% \Rightarrow 30\mathrm{g}\) - S: \(30 \% \Rightarrow 30\mathrm{g}\)

Step 3: 3. Convert the mass of each reactant to moles

Next, we need to convert the mass of each reactant to moles using molar masses. - Moles of KNO₃: \(\frac{40\mathrm{g}}{101.1\mathrm{g/mol}} = 0.395\mathrm{moles}\) - Moles of C: \(\frac{30\mathrm{g}}{12.01\mathrm{g/mol}} = 2.50\mathrm{moles}\) - Moles of S: \(\frac{30\mathrm{g}}{32.07\mathrm{g/mol}} = 0.936\mathrm{moles}\)

Step 4: 4. Determine the limiting reagent

Now we must determine the limiting reagent. To do this, we must compare the ratios of the moles of the reactants to their stoichiometric coefficients: - KNO₃: \(\frac{0.395\mathrm{moles}}{2} = 0.1975\) - C: \(\frac{2.50\mathrm{moles}}{3} = 0.8333\) - S: \(\frac{0.936\mathrm{moles}}{1} = 0.936\) The smallest ratio value corresponds to the limiting reagent, which is KNO₃.

Step 5: 5. Calculate the amount of each excess reagent left over after the reaction

We can now calculate the amount of excess reagent left over for carbon and sulfur after the reaction. We will use stoichiometry and the limiting reagent KNO₃ to do this: - Excess Carbon: - Moles of C reacted: \(3 \times 0.395\mathrm{moles\, KNO_{3}} = 1.185\mathrm{moles\, C}\) - Moles of C left over: \(2.50\mathrm{moles} - 1.185\mathrm{moles} = 1.315\mathrm{moles\, C}\) - Mass of C left over: \(1.315\mathrm{moles} \times 12.01\mathrm{g/mol} = 15.8\mathrm{g}\) - Excess Sulfur: - Moles of S reacted: \(1 \times 0.395\mathrm{moles\, KNO_{3}} = 0.395\mathrm{moles\, S}\) - Moles of S left over: \(0.936\mathrm{moles} - 0.395\mathrm{moles} = 0.541\mathrm{moles\, S}\) - Mass of S left over: \(0.541\mathrm{moles} \times 32.07\mathrm{g/mol} = 17.3\mathrm{g}\)

Step 6: Answer#a. The limiting reagent is potassium nitrate (KNO₃).

b. After the explosion, there will be \(15.8\mathrm{g}\) of carbon and \(17.3\mathrm{g}\) of sulfur left over.

Key Concepts

Limiting Reagent

When dealing with chemical reactions, the concept of the limiting reagent is essential. A limiting reagent is the reactant that gets completely used up first and determines the amount of products formed. Once the limiting reagent is consumed, the reaction stops.
In our black powder example, we have potassium nitrate, carbon, and sulfur reacting together. To identify the limiting reagent, we calculate the moles of each reactant and then compare these moles to the coefficients in the balanced chemical equation.
In this case, potassium nitrate (KNO₃) is the limiting reagent as it has the smallest ratio of moles to its stoichiometric coefficient. When working through such problems, always remember to use the smallest ratio to identify your limiting reagent.

Stoichiometry

Stoichiometry is the mathematical relationship between the amounts of reactants and products in a chemical reaction. It helps us predict how much product we can get from a given quantity of reactant.
In the black powder exercise, stoichiometry allows us to understand the relationship between potassium nitrate, carbon, and sulfur, leading to various gaseous products and potassium sulfite. By using the balanced chemical equation, we can derive the stoichiometric coefficients that tell us how much of each reactant is required.
The chemical reaction for black powder, \(2\text{ KNO}_3 + 3\text{ C} + \text{ S} \rightarrow 3\text{ CO} + \text{ CO}_2 + \text{ K}_2\text{SO}_3 + \frac{1}{2}\text{ N}_2\),shows these relationships clearly. The coefficients tell us that for every 2 moles of potassium nitrate, we need 3 moles of carbon and 1 mole of sulfur.

Chemical Reactions

Chemical reactions are processes where reactants transform into products. This transformation involves breaking bonds in the reactants and forming new bonds to create the products.
For black powder, the reaction is a classic combination of reactants becoming gaseous products, which provides explosive force. Key gaseous products from this reaction include carbon monoxide, carbon dioxide, nitrogen, and solid potassium sulfite.
Understanding the reaction involves writing a balanced chemical equation. The balanced equation reflects the law of conservation of mass, ensuring that the mass of the reactants equals the mass of the products, demonstrating that atoms are neither created nor destroyed.

Molar Mass Calculations

To successfully solve chemistry problems, one must be proficient in molar mass calculations. Molar mass is the mass of one mole of a substance, usually in grams per mole (g/mol). It's a crucial figure for converting between the mass of a substance and the amount in moles, linking mass and quantity.
In calculating the limiting reagent for black powder, knowing the molar masses of potassium nitrate (KNO₃), carbon (C), and sulfur (S), is key.
For instance, the molar mass of KNO₃ is approximately 101.1 g/mol, carbon is 12.01 g/mol, and sulfur is 32.07 g/mol. By dividing the mass of each component by its molar mass, we can find the number of moles, which is crucial for further stoichiometric calculations.
Mastering molar mass helps streamline the process of converting amounts in chemistry problems, leading to more accurate results.