Question

Ethane burns in air to give \(\mathrm{H}_{2} \mathrm{O}\) and \(\mathrm{CO}_{2}\) $$ 2 \mathrm{C}_{2} \mathrm{H}_{6}(\mathrm{g})+7 \mathrm{O}_{2}(\mathrm{g}) \rightarrow 4 \mathrm{CO}_{2}(\mathrm{g})+6 \mathrm{H}_{2} \mathrm{O}(\mathrm{g}) $$ What volume of \(\mathrm{O}_{2}(\mathrm{L})\) is required for complete reaction with \(5.2 \mathrm{L}\) of \(\mathrm{C}_{2} \mathrm{H}_{6}\) ? What volume of \(\mathrm{H}_{2} \mathrm{O}\) vapor (L) is produced? Assume all gases are measured at the same temperature and pressure.

Short answer

18.2 L of \(\mathrm{O}_{2}\) is required, and 15.6 L of \(\mathrm{H}_{2} \mathrm{O}\) vapor is produced.

Step-by-step solution

Identify the Reaction

The balanced chemical reaction is given as \(2 \mathrm{C}_{2} \mathrm{H}_{6}(\mathrm{g})+7 \mathrm{O}_{2}(\mathrm{g}) \rightarrow 4 \mathrm{CO}_{2}(\mathrm{g})+6 \mathrm{H}_{2} \mathrm{O}(\mathrm{g})\). This tells us that 2 moles of ethane react with 7 moles of oxygen to produce 4 moles of carbon dioxide and 6 moles of water.

Use the Mole Ratio

From the balanced equation, the mole ratio between \(\mathrm{C}_{2} \mathrm{H}_{6}\) and \(\mathrm{O}_{2}\) is 2:7. Therefore, for every 2 moles (or liters, under the same temperature and pressure) of ethane, 7 moles (or liters) of oxygen are required.

Calculate O2 Volume Required

Using the ratio, to find the volume of \(\mathrm{O}_{2}\) needed for 5.2 L of \(\mathrm{C}_{2} \mathrm{H}_{6}\): \(\text{Volume of } \mathrm{O}_{2} = \frac{7}{2} \times 5.2\text{ L}\ = 18.2\text{ L}\).

Use the Mole Ratio for H2O

The mole ratio between \(\mathrm{C}_{2} \mathrm{H}_{6}\) and \(\mathrm{H}_{2} \mathrm{O}\) is 2:6, or 1:3. For every 1 mole or liter of ethane burned, 3 moles or liters of water vapor are produced.

Calculate H2O Volume Produced

To find the volume of \(\mathrm{H}_{2} \mathrm{O}\) produced from 5.2 L of \(\mathrm{C}_{2} \mathrm{H}_{6}\): \(\text{Volume of } \mathrm{H}_{2} \mathrm{O} = 3 \times 5.2\text{ L}\ = 15.6\text{ L}\).

Key concepts

Gas Laws

In chemistry, the behavior of gases is often described using various gas laws. These laws relate the volume, pressure, and temperature of a gas with the amount of gas present, usually measured in moles. In the problem at hand, we assume that all gases are measured under the same conditions of temperature and pressure. This simplifies the problem significantly, allowing us to make direct comparisons between volumes of gases.

Because of these consistent conditions, we can use the volumes directly in our calculations involving stoichiometry. This is thanks to Avogadro's Law, which states that equal volumes of gases at the same temperature and pressure contain an equal number of moles. Therefore, the volume of a gas is directly proportional to the number of moles, as long as temperature and pressure remain constant.

• Allows for mole-to-volume conversions without additional calculations.
• Simplifies stoichiometric problems significantly under constant conditions.

Balanced Chemical Equation

A balanced chemical equation is like a recipe for a chemical reaction. It shows the reactants, the products, and the proportions in which they react or are produced. In the case of our ethane combustion reaction, the balanced equation is:\[ 2 \ \mathrm{C}_{2} \ \mathrm{H}_{6} (g) + 7 \ \mathrm{O}_{2} (g) \rightarrow 4 \ \mathrm{CO}_{2} (g) + 6 \ \mathrm{H}_{2} \mathrm{O} (g) \]This equation tells us that:

• 2 molecules (or moles) of ethane react with 7 molecules (or moles) of oxygen.
• The reaction produces 4 molecules (or moles) of carbon dioxide and 6 molecules (or moles) of water.

A balanced equation ensures the law of conservation of mass is satisfied, meaning the mass of reactants equals the mass of products.

By balancing the chemical equation, we ensure that the number of atoms of each element is the same on both sides of the equation. This balance allows us to use the equation quantitatively to determine the amounts of reactants needed or products formed.

• Makes it possible to calculate precise amounts in reactions.
• Foundational for understanding chemical reactions and stoichiometry.

Mole Ratio

The mole ratio is a crucial concept in stoichiometry, making it possible to relate quantities of reactants and products in a balanced chemical equation. It is simply the ratio of moles of one substance to the moles of another, based on the coefficients in the balanced equation.

In our ethane combustion example, the balanced equation gives us a mole ratio of:

• Ethane (\( \mathrm{C}_{2} \mathrm{H}_{6} \)) to Oxygen (\( \mathrm{O}_{2} \)) is 2:7.
• Ethane (\( \mathrm{C}_{2} \mathrm{H}_{6} \)) to Water (\( \mathrm{H}_{2} \mathrm{O} \)) is 2:6, which simplifies to 1:3.

This means for every 2 moles of ethane, 7 moles of oxygen are required, and 3 moles of water vapor are produced per mole of ethane.

Using the mole ratio, you can calculate the volume of gases involved in the reaction if the conditions of temperature and pressure are constant.
The ability to calculate these conversions allows for precise predictions about the quantities of materials consumed and generated in chemical processes.

• Essential for solving stoichiometric problems accurately.
• Allows conversion between moles and volumes in gas reactions under constant conditions.