Problem 18 - Chemistry and Chemical Reactivity

Question

Ethane, \(\mathrm{C}_{2} \mathrm{H}_{6},\) burns in oxygen. (a) What are the products of the reaction? (b) Write the balanced equation for the reaction. (c) What mass of \(\mathrm{O}_{2}\), in grams, is required for complete combustion of 13.6 of ethane? (d) What is the total mass of products expected from the combustion of \(13.6 \mathrm{g}\) of ethane?

Step-by-Step Solution

Answer

(a) Carbon dioxide and water; (b) \(\mathrm{2C}_{2} \mathrm{H}_{6} + 7\mathrm{O}_{2} \rightarrow 4 \mathrm{CO}_{2} + 6 \mathrm{H}_{2}\mathrm{O}\); (c) 50.624 g of \(\mathrm{O}_{2}\); (d) 64.224 g of products.

Step 1: Identify the Products of Combustion

When ethane, \( \mathrm{C}_{2} \mathrm{H}_{6} \), burns in oxygen, it produces carbon dioxide \( \mathrm{CO}_2 \) and water \( \mathrm{H}_2\mathrm{O} \) as the products of the combustion reaction. Thus, the products are carbon dioxide and water.

Step 2: Write the Unbalanced Chemical Equation

The unbalanced chemical equation for the combustion of ethane in oxygen can be written as: \[ \mathrm{C}_{2} \mathrm{H}_{6} + \mathrm{O}_{2} \rightarrow \mathrm{CO}_{2} + \mathrm{H}_{2}\mathrm{O} \]

Step 3: Balance the Chemical Equation

To balance the equation, adjust the coefficients to conserve the number of each type of atom on both sides:- C atoms: \( \mathrm{C}_{2} \) requires 2 \( \mathrm{CO}_{2} \), so change to \( 2 \mathrm{CO}_{2} \).- H atoms: \( \mathrm{H}_{6} \) requires 3 \( \mathrm{H}_2\mathrm{O} \), so change to \( 3 \mathrm{H}_2\mathrm{O} \).- O atoms: Needs 7 \( \mathrm{O}_{2} \), since \( 2 \times 2 + 3 = 7 \) O atoms.The balanced equation becomes: \[ \mathrm{2C}_{2} \mathrm{H}_{6} + 7\mathrm{O}_{2} \rightarrow 4 \mathrm{CO}_{2} + 6 \mathrm{H}_{2}\mathrm{O} \]

Step 4: Calculate Moles of Ethane

Given 13.6 g of ethane, first find the number of moles using the molar mass of ethane (\( \mathrm{C}_{2} \mathrm{H}_{6} = 30.07 \ \mathrm{g/mol} \)):\( \text{Moles of } \mathrm{C}_{2} \mathrm{H}_{6} = \frac{13.6 \ \mathrm{g}}{30.07 \ \mathrm{g/mol}} = 0.452 \ \mathrm{mol} \).

Step 5: Use Stoichiometry to Find Oxygen Mass Required

According to the balanced equation, 1 mole of \( \mathrm{C}_{2} \mathrm{H}_{6} \) requires 3.5 moles of \( \mathrm{O}_{2} \). Thus, 0.452 moles of \( \mathrm{C}_{2} \mathrm{H}_{6} \) requires \( 0.452 \times 3.5 = 1.582 \ \mathrm{mol} \) of \( \mathrm{O}_{2} \).Then find the mass of \( \mathrm{O}_{2} \) using its molar mass (32.00 g/mol):\( \text{Mass of } \mathrm{O}_{2} = 1.582 \times 32.00 = 50.624 \ \mathrm{g} \).

Step 6: Calculate Total Mass of Products

The law of conservation of mass states that total mass of products equals total mass of reactants. The total mass of reactants is 13.6 g of \( \mathrm{C}_{2} \mathrm{H}_{6} \) plus mass of \( \mathrm{O}_{2} \) (50.624 g), so the total mass is 13.6 + 50.624 = 64.224 g.Hence, the total mass of products expected is \( 64.224 \ \mathrm{g} \).

Key Concepts

Chemical Equation Balancing

Balancing chemical equations is essential to understanding chemical reactions. When ethane (\( \mathrm{C}_{2}\mathrm{H}_{6} \)) combusts in oxygen (\( \mathrm{O}_{2} \)), it forms carbon dioxide (\( \mathrm{CO}_{2} \)) and water (\( \mathrm{H}_{2}\mathrm{O} \)). To balance this equation, one must adjust the coefficients so that the number of each type of atom is equal on both the reactants and products side. This is a critical step because it aligns with the principle of conservation of mass.
In our example, you start with the unbalanced equation:\[ \mathrm{C}_{2} \mathrm{H}_{6} + \mathrm{O}_{2} \rightarrow \mathrm{CO}_{2} + \mathrm{H}_{2}\mathrm{O} \]The balancing process involves looking at different atoms one at a time:

• For carbon: Since there are 2 carbons in \( \mathrm{C}_{2}\mathrm{H}_{6} \), you need 2 \( \mathrm{CO}_{2} \).
• For hydrogen: 6 hydrogens in \( \mathrm{C}_{2}\mathrm{H}_{6} \) require 3 \( \mathrm{H}_{2}\mathrm{O} \).
• Finally, balance the oxygen: Total 7 oxygen molecules are needed from \( \mathrm{O}_{2} \).

The balanced equation becomes:\[ \mathrm{2C}_{2} \mathrm{H}_{6} + 7\mathrm{O}_{2} \rightarrow 4 \mathrm{CO}_{2} + 6 \mathrm{H}_{2}\mathrm{O} \] This balancing ensures the law of conservation of mass is adhered to.

Stoichiometry

Stoichiometry is used to calculate quantities in chemical reactions. This involves using the balanced chemical equation to understand the relationships between reactants and products in a reaction.
In the case of ethane combustion, stoichiometry helps determine how much oxygen is needed or how much product is formed. For instance, from the balanced equation \( \mathrm{2C}_{2}\mathrm{H}_{6} + 7\mathrm{O}_{2} \rightarrow 4 \mathrm{CO}_{2} + 6 \mathrm{H}_{2}\mathrm{O} \), it tells us:

• Two moles of \( \mathrm{C}_{2}\mathrm{H}_{6} \) react with seven moles of \( \mathrm{O}_{2} \).
• Produce four moles of \( \mathrm{CO}_{2} \) and six moles of \( \mathrm{H}_{2}\mathrm{O} \).

Since stoichiometry is about these relationships, it means that if you have a particular quantity of one substance, you can calculate the quantity of another using mole ratios. In this problem, if you have 13.6 g of ethane, stoichiometry allows you to determine the mass of oxygen needed and the mass of carbon dioxide and water produced.

Conservation of Mass

The conservation of mass principle states that mass in a closed system will remain constant. This means that the mass of the reactants will equal the mass of the products in a chemical reaction.
In this exercise, for the combustion of ethane (\( \mathrm{C}_{2}\mathrm{H}_{6} \)), the total mass of the reactants, ethane, and oxygen, should equal the total mass of the products, carbon dioxide, and water.
When 13.6 g of ethane combusts, the calculation takes into account:

• 13.6 g of ethane used.
• The calculated 50.624 g of oxygen required to combust all the ethane.

Thus, the total mass before and after the reaction remains constant at 64.224 g, confirming the conservation of mass. This principle is crucial in confirming that a chemical equation is properly balanced.

Molar Mass Calculations

Understanding molar mass calculations is key to converting between grams and moles, which is essential in stoichiometry.
Molar mass is the mass of one mole of a substance, measured in grams per mole (\( \mathrm{g/mol} \)). For ethane (\( \mathrm{C}_{2}\mathrm{H}_{6} \)), calculate its molar mass by summing the atomic masses of its components:

• 2 carbon atoms, each \( 12.01 \ \mathrm{g/mol} \)
• 6 hydrogen atoms, each \( 1.01 \ \mathrm{g/mol} \)

Giving \( 12.01 \times 2 + 1.01 \times 6 = 30.07 \ \mathrm{g/mol} \).
To find the moles of ethane in 13.6 g, use the formula:\[\text{Moles of ethane} = \frac{\text{Mass of ethane}}{\text{Molar mass of ethane}} = \frac{13.6 \ \mathrm{g}}{30.07 \ \mathrm{g/mol}} \approx 0.452 \ \mathrm{mol}\]These calculations enable you to compute how much ethane reacts with a specific amount of oxygen in a balanced chemical reaction.