Problem 103 - Chemistry The Central Science

Question

When a mixture of \(10.0 \mathrm{~g}\) of acetylene \(\left(\mathrm{C}_{2} \mathrm{H}_{2}\right) ;\) and \(10.0 \mathrm{~g}\) of oxygen \(\left(\mathrm{O}_{2}\right)\) is ignited, the resultant combustion reaction produces \(\mathrm{CO}_{2}\) and \(\mathrm{H}_{2} \mathrm{O}\) (a) Write the balanced chemical equation for this reaction. (b) Which is the limiting reactant? (c) How many grams of \(\mathrm{C}_{2} \mathrm{H}_{2}, \mathrm{O}_{2}, \mathrm{CO}_{2},\) and \(\mathrm{H}_{2} \mathrm{O}\) are present after the reaction is complete?

Step-by-Step Solution

Answer

After the combustion reaction is complete: - Mass of remaining C2H2: 6.71 g - Mass of remaining O2: 0 g - Mass of generated CO2: 11.04 g - Mass of generated H2O: 2.27 g

Step 1: Write the balanced chemical equation

The balanced chemical equation for the combustion of acetylene (C2H2) in the presence of oxygen (O2) forming carbon dioxide (CO2) and water (H2O) is: \[2 C_{2}H_{2} + 5 O_{2} \rightarrow 4 CO_{2} + 2 H_{2}O\]

Step 2: Calculate moles of reactants

To determine the limiting reactant, we will first calculate the moles of each reactant present: Given that there are 10 grams of each C2H2 and O2, Moles of C2H2 = \(\frac{10.0 \ \text{g}}{26.04 \ \text{g/mol}} = 0.384 \ \text{moles}\) Moles of O2 = \(\frac{10.0 \ \text{g}}{32.00 \ \text{g/mol}} = 0.313 \ \text{moles}\)

Step 3: Determine the limiting reactant

Now, we will check which reactant is the limiting reactant: Dividing the moles of each reactant by its respective stoichiometric coefficient: C2H2: \(\frac{0.384 \ \text{moles}}{2} = 0.192\) O2: \(\frac{0.313 \ \text{moles}}{5} = 0.063\) Since the value for O2 (0.063) is smaller than the value for C2H2 (0.192), oxygen is the limiting reactant.

Step 4: Determine the mass of each substance after the reaction

Using the limiting reactant, we can calculate the mass of each substance present after the reaction: (a) Mass of remaining C2H2: Since 5 moles of O2 react with 2 moles of C2H2, \(\ 0.313\ \text{moles} \times \frac{2 \ \text{moles} \ C2H2}{5 \ \text{moles} \ O2} = 0.126\ \text{moles} \ \text{of consumed C2H2}\) Remaining C2H2 = Initial C2H2 - Consumed C2H2 = \(0.384 \ \text{moles} - 0.126 \ \text{moles} = 0.258 \ \text{moles}\) Mass of remaining C2H2 = \(0.258 \ \text{moles} \times 26.04 \ \text{g/mol} = 6.71\ \text{g}\) (b) Mass of remaining O2: Since O2 is the limiting reactant, it must be fully consumed, so the mass of remaining O2 is \(0\ \text{g}\). (c) Mass of generated CO2: Given 2 moles of C2H2 react with 5 moles of O2 to form 4 moles of CO2, \(\ 0.313\ \text{moles} \times \frac{4 \ \text{moles} \ CO2}{5 \ \text{moles} \ O2} = 0.251\ \text{moles} \ \text{of CO2}\) Mass of generated CO2 = \(0.251\ \text{moles} \times 44.01\ \text{g/mol} = 11.04\ \text{g}\) (d) Mass of generated H2O: Given 2 moles of C2H2 react with 5 moles of O2 to form 2 moles of H2O, \(\ 0.313\ \text{moles} \times \frac{2 \ \text{moles} \ H2O}{5 \ \text{moles} \ O2} = 0.126\ \text{moles} \ \text{of H2O}\) Mass of generated H2O = \(0.126 \ \text{moles} \times 18.02\ \text{g/mol} = 2.27\ \text{g}\)

Step 5: Final answer

After the combustion reaction is complete: - Mass of remaining C2H2: 6.71 g - Mass of remaining O2: 0 g - Mass of generated CO2: 11.04 g - Mass of generated H2O: 2.27 g

Key Concepts

Balanced Chemical Equations

A balanced chemical equation ensures the same number of each type of atom on both sides of the equation, representing the conservation of mass. In the reaction given, acetylene \(\mathrm{C_{2}H_{2}}\) burns in oxygen \(\mathrm{O_{2}}\) to form carbon dioxide \(\mathrm{CO_{2}}\) and water \(\mathrm{H_{2}O}\). To balance this, we ensure that each element has the same quantity on both sides of the equation. For instance, the equation \[2 \mathrm{C_{2}H_{2}} + 5 \mathrm{O_{2}} \rightarrow 4 \mathrm{CO_{2}} + 2 \mathrm{H_{2}O}\] ensures that we have 4 carbon atoms, 4 hydrogen atoms, and 10 oxygen atoms on each side.

Here's how to balance a chemical equation step by step:

• Count the number of atoms for each element in the reactants and products.
• Adjust the coefficients, which are the numbers in front of molecules, to balance the atoms for each element.
• Repeat the counting and adjusting until all elements are balanced.

Balanced equations are fundamental for understanding the proportion of reactants and products involved in chemical reactions.

Stoichiometry

Stoichiometry is the quantitative relationship between the number of moles of reactants and products in a chemical reaction. This concept uses the balanced chemical equation to predict the amounts of products generated based on the reactants consumed. For our combustion reaction, stoichiometry helps us calculate how much \(\mathrm{CO_{2}}\) and \(\mathrm{H_{2}O}\) we produce from given amounts of \(\mathrm{C_{2}H_{2}}\) and \(\mathrm{O_{2}}\).

Let's break down how stoichiometry is applied:

• First, determine the molar ratio of reactants to products from the balanced equation. In this case, 2 moles of \(\mathrm{C_{2}H_{2}}\) react with 5 moles of \(\mathrm{O_{2}}\) to produce 4 moles of \(\mathrm{CO_{2}}\) and 2 moles of \(\mathrm{H_{2}O}\).


• Next, calculate the moles of \(\mathrm{C_{2}H_{2}}\) and \(\mathrm{O_{2}}\) using their masses and molar masses.


• Use these moles and the molar ratio to find out how much product you should expect.

Stoichiometry provides the foundation for reactors, predicting yields, and ensuring that industrial processes have the correct proportions of chemical components.

Limiting Reactant

The limiting reactant is the substance that is completely consumed in a chemical reaction, determining the total amount of product formed. In any reaction, one reactant typically runs out before others, which constrains the production of products. For our example of \(\mathrm{C_{2}H_{2}}\) and \(\mathrm{O_{2}}\), we determine which is the limiting reactant by comparing the mole ratio from the balanced equation.

Here's how to identify the limiting reactant:

• Calculate the moles of each reactant.
• Divide the moles of each reactant by its stoichiometric coefficient from the balanced equation.
• The reactant with the smallest quotient is the limiting reactant.

Using these steps, oxygen \(\mathrm{O_{2}}\) is found to be the limiting reactant in the given reaction. The limiting reactant concept is crucial for optimizing reactions and reducing wastage in chemical synthesis.

Combustion Reactions

Combustion reactions are a type of chemical reaction where a substance reacts rapidly with oxygen, releasing energy in the form of heat and light. Such reactions are exothermic and typically produce carbon dioxide and water as products when organic compounds are involved. The combustion of acetylene \(\mathrm{C_{2}H_{2}}\), as in our exercise, is a classic example:

- \(\mathrm{C_{2}H_{2}}\) burns in the presence of \(\mathrm{O_{2}}\) to form \(\mathrm{CO_{2}}\) and \(\mathrm{H_{2}O}\).

Combustion reactions are essential in many applications, such as:

• Power generation in engines and turbines.
• Heating homes and cooking.
• Fireworks and explosives where controlled combustion is used for effect.

Understanding combustion reactions helps in managing energy resources more efficiently and designing safer industrial processes.