Problem 77 - Chemistry: Principles and Reactions

Question

Oxyacetylene torches used for welding reach temperatures near \(2000^{\circ} \mathrm{C}\). The reaction involved in the combustion of acetylene is $$ 2 \mathrm{C}_{2} \mathrm{H}_{2}(g)+5 \mathrm{O}_{2}(g) \longrightarrow 4 \mathrm{CO}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(g) $$ (a) Starting with \(175 \mathrm{~g}\) of both acetylene and oxygen, what is the theoretical yield, in grams, of carbon dioxide? (b) If \(68.5 \mathrm{~L}(d=1.85 \mathrm{~g} / \mathrm{L})\) of carbon dioxide is produced, what is the percent yield at the same conditions of temperature and pressure? (c) How much of the reactant in excess is unused? (Assume 100\% yield.)

Step-by-Step Solution

Answer

Answer: The theoretical yield of CO2 is 192.56 g, the percent yield of CO2 is 65.8%, and the amount of unreacted acetylene is 117.93 g.

Step 1: (a) Calculate the theoretical yield of CO2

: Step 1: Convert grams of reactants to moles We are given 175 g of acetylene (C2H2) and 175 g of oxygen (O2). We first need to calculate their respective moles using their molar masses. Molar mass of C2H2: 2(12.01) + 2(1.01) = 26.04 g/mol Moles of C2H2: (175 g) / (26.04 g/mol) = 6.72 moles Molar mass of O2: 2(16.00) = 32.00 g/mol Moles of O2: (175 g) / (32.00 g/mol) = 5.47 moles Step 2: Identify the limiting reactant The balanced chemical equation is: 2 C2H2 + 5 O2 -> 4 CO2 + 2 H2O From the balanced equation, 2 moles of acetylene react with 5 moles of oxygen. To determine the limiting reactant, we can calculate the mole ratios: Mole ratio of C2H2: 6.72 / 2 = 3.36 Mole ratio of O2: 5.47 / 5 = 1.09 Since the mole ratio of O2 is smaller, oxygen is the limiting reactant. Step 3: Calculate the theoretical yield of CO2 in grams Using the stoichiometry of the balanced equation, 5 moles of O2 produce 4 moles of CO2. Moles of CO2 produced: 5.47 moles O2 * (4 moles CO2 / 5 moles O2) = 4.375 moles CO2 Molar mass of CO2: (12.01) + 2(16.00) = 44.01 g/mol Theoretical yield of CO2: (4.375 moles CO2) * (44.01 g/mol) = 192.56 g The theoretical yield of CO2 is 192.56 g.

Step 2: (b) Calculate the percent yield of CO2

: Step 4: Convert the volume of CO2 produced to grams The volume of CO2 produced is given as 68.5 L, with a density of 1.85 g/L. Mass of CO2 produced: 68.5 L * 1.85 g/L = 126.725 g Step 5: Calculate the percent yield Percent yield = (Actual yield / Theoretical yield) * 100 Percent yield = (126.725 g / 192.56 g) * 100 = 65.8 % The percent yield of CO2 is 65.8%.

Step 3: (c) Calculate the amount of unreacted reactant in excess

: Step 6: Calculate the amount of unreacted acetylene Since all oxygen is consumed (as it is the limiting reactant), we will calculate the remaining amount of acetylene using the stoichiometry of the balanced equation. The initial moles of acetylene were 6.72 moles. The stoichiometry states that 5 moles O2 consume 2 moles C2H2. Moles of C2H2 consumed: 5.47 moles O2 * (2 moles C2H2 / 5 moles O2) = 2.188 moles C2H2 Moles of unreacted C2H2: 6.72 moles - 2.188 moles = 4.532 moles Mass of unreacted C2H2: 4.532 moles * 26.04 g/mol = 117.93 g The amount of unused acetylene is 117.93 g.

Key Concepts

Stoichiometry

Stoichiometry is at the heart of chemical reactions. It involves the calculation of the quantities of reactants and products involved in a chemical reaction. Understanding stoichiometry is crucial because it lays the foundation for predicting the outcomes of reactions and for the conservation of mass in chemical processes. The core principle is that elements are neither created nor destroyed in a chemical reaction, meaning the amount of each element must be the same in both the reactants and the products.

To perform stoichiometric calculations, we first need a balanced chemical equation. This equation tells us the ratios in which reactants combine and products form. For example, in the combustion of acetylene, the balanced equation is: \[2 \text{C}_{2}\text{H}_{2}(g) + 5 \text{O}_{2}(g) \rightarrow 4 \text{CO}_{2}(g) + 2 \text{H}_{2}\text{O}(g)\]. Each component of the reaction, from acetylene to oxygen to carbon dioxide and water, has a corresponding 'mole ratio' that governs how much of one substance reacts with or produces another.

By using the mole concept, which relates the mass of a substance to the number of particles it contains, we can convert between mass and moles and apply the ratios from the balanced equation to determine how much product can be obtained from given amounts of reactants. This approach is an invaluable tool for ensuring that reactions are carried out with the appropriate amounts of materials, thereby avoiding waste and ensuring the desired outcome.

Theoretical Yield Calculation

The theoretical yield is the maximum amount of product that can be produced in a chemical reaction if everything went perfectly, meaning all the limiting reactant is completely used up without any losses. The calculation of theoretical yield is a key step in the process of stoichiometry and particularly relevant for industries that produce chemicals, where maximizing efficiency and productivity is crucial.

To calculate the theoretical yield, one must:

1. Identify the limiting reactant: the reactant that will be completely consumed first in a chemical reaction, limiting the amount of product that can be formed.
2. Use stoichiometry to determine the amount of product that can be formed from the limiting reactant.
3. Convert the number of moles of product to mass if required.

Using the acetylene combustion example, oxygen was determined to be the limiting reactant. Knowing the amount of the limiting reactant, and the balanced equation, we can then calculate how many moles of carbon dioxide should theoretically be produced and, subsequently, its mass. This provides a benchmark for what is possible under ideal conditions, which is essential for comparing with the actual yield obtained from a real-world reaction.

Percent Yield

Once a reaction has been completed, it's necessary to evaluate the efficiency of the process, and percent yield is a measure of this efficiency. Percent yield is the ratio of the actual yield (the amount of product actually obtained from the reaction) to the theoretical yield (the maximum possible amount of product), multiplied by 100 to convert the ratio to a percentage.

To find percent yield, we use the formula: \[\text{Percent Yield} = \left(\frac{\text{Actual Yield}}{\text{Theoretical Yield}}\right) \times 100\%\]

By comparing the actual yield to the theoretical yield, we can gain insights into the practical limitations of a chemical process, such as losses in handling, side-reactions that consume some reactants, or incomplete reactions. The percent yield can never exceed 100%. A yield less than 100% is normal and expected, due to practical considerations. In the combustion of acetylene exercise, after calculating that 192.56 grams of carbon dioxide is the theoretical yield, and that only 126.725 grams were actually obtained, applying the percent yield formula indicates that the process was 65.8% efficient. It's important to note that the actual yield may vary due to different experimental conditions or measurement inaccuracies. Thus, the percent yield provides a quantitative method for comparing the actual outcome of a reaction to the ideal based on stoichiometric predictions.