Question

Liquid benzene \(\left(\mathrm{C}_{6} \mathrm{H}_{6}\right)\) burns in oxygen according to \(2 \mathrm{C}_{6} \mathrm{H}_{6}(\mathrm{l})+15 \mathrm{O}_{2}(\mathrm{~g}) \rightarrow 12 \mathrm{CO}_{2}(\mathrm{~g})+6 \mathrm{H}_{2} \mathrm{O}(\mathrm{g})\) How many litres of \(\mathrm{O}_{2}\) at STP are needed to complete the combustion of \(39 \mathrm{~g}\) of liquid benzene? (Mol. wt. of \(\mathrm{O}_{2}=32\), \(\left.\mathrm{C}_{6} \mathrm{H}_{6}=78\right)\) a. \(74 \mathrm{~L}\) b. \(11.2 \mathrm{~L}\) c. \(22.4 \mathrm{~L}\) d. \(84 \mathrm{~L}\)

Short answer

d. 84 L

Step-by-step solution

Calculate Moles of Benzene

To find out how many moles of benzene (\(\text{C}_6\text{H}_6\)) are in 39 grams, use the formula for moles: \( \text{moles} = \frac{\text{mass}}{\text{molar mass}} \). The molar mass of benzene is 78 g/mol. So, \( \text{moles of } \text{C}_6\text{H}_6 = \frac{39}{78} = 0.5 \text{ moles} \).

Determine Moles of Oxygen Required

From the balanced chemical equation, \(2 \text{C}_6\text{H}_6 + 15 \text{O}_2 \rightarrow 12 \text{CO}_2 + 6 \text{H}_2\text{O}\), it is clear that 2 moles of benzene require 15 moles of oxygen gas (\(\text{O}_2\)). Therefore, for 0.5 moles of benzene, \(0.5 \times \frac{15}{2} = 3.75 \text{ moles of } \text{O}_2\) are needed.

Convert Moles of Oxygen to Litres

At Standard Temperature and Pressure (STP), 1 mole of any gas occupies 22.4 L. Therefore, to find the volume of oxygen gas required, multiply the moles of \(\text{O}_2\) by 22.4 L/mole: \(3.75 \times 22.4 = 84 \text{ L}\) of \(\text{O}_2\).

Key concepts

Understanding Stoichiometry

Stoichiometry is the section of chemistry that involves calculating the relationships between reactants and products in a chemical reaction. It helps in predicting the amount of substances consumed and produced in a given reaction.
To begin with, mole ratios derived from balanced chemical equations are key. They serve as conversion factors that link amounts of reactants and products.

For example, in the combustion of benzene given by the equation:\[ 2 \text{C}_6\text{H}_6 + 15 \text{O}_2 \rightarrow 12 \text{CO}_2 + 6 \text{H}_2\text{O} \]The stoichiometry tells us that:

• 2 moles of benzene react with 15 moles of oxygen.
• From this, you can derive that each mole of benzene will need 7.5 moles of oxygen.

This concept is crucial when calculating the reactant requirements or the product produced. When you know the quantity of one substance, stoichiometry allows you to determine the quantity of another substance involved in the reaction.

Molar Volume at STP

One of the key concepts in gas stoichiometry is the molar volume at STP, which helps to convert moles of a gas to liters.
STP stands for Standard Temperature and Pressure, typically recognized as 0°C (273.15 K) and 1 atm pressure. Under these conditions, any ideal gas occupies 22.4 liters per mole according to Avogadro's law.

This principle helps in calculations involving gases in a chemical reaction:

• To convert moles of gas to a volume at STP, multiply by 22.4 L/mole.
• In the case of oxygen gas in the combustion of benzene, 3.75 moles of \(\text{O}_2\) equates to \(3.75 \times 22.4 = 84\) liters.

Using the molar volume at STP simplifies the process of calculating the volume of gases that participate in chemical reactions, making it a fundamental aspect of quantitative chemistry.

Balancing Chemical Equations

A balanced chemical equation is essential in a chemical reaction as it ensures that the same number of each type of atom is present on both sides of the equation.
This aligns with the law of conservation of mass, stating that mass cannot be created or destroyed in a chemical reaction.

Let's look at the balanced equation for the combustion of benzene:

• The equation \(2 \text{C}_6\text{H}_6 + 15 \text{O}_2 \rightarrow 12 \text{CO}_2 + 6 \text{H}_2\text{O}\) shows 2 moles of \(\text{C}_6\text{H}_6\) react with 15 moles of \(\text{O}_2\).
• This results in 12 moles of carbon dioxide \(\text{CO}_2\) and 6 moles of water \(\text{H}_2\text{O}\).
• Balancing equations is essential because it provides the exact quantities needed to react completely without any leftover reactants.

Balancing chemical equations confirms that chemical reactions observe the conservation of atom counts, ensuring no atoms are lost or gained, only rearranged.