Question

How many liters of \(\mathrm{O}_{2}\) at STP are required to burn \(3.77 \mathrm{~g}\) of butane from a disposable lighter? \(2 \mathrm{C}_{4} \mathrm{H}_{10}(\mathrm{~g})+13 \mathrm{O}_{2}(\mathrm{~g}) \rightarrow 8 \mathrm{CO}_{2}(\mathrm{~g})+10 \mathrm{H}_{2} \mathrm{O}(\ell)\)

Short answer

9.45 liters of \( \mathrm{O}_{2} \) are needed.

Step-by-step solution

Calculate Moles of Butane

First, we need to calculate the number of moles of butane (\( \mathrm{C}_4\mathrm{H}_{10} \)) in 3.77 g. The molar mass of butane (\( \mathrm{C}_4\mathrm{H}_{10} \)) is 58.12 g/mol.\[\text{Moles of } \mathrm{C}_4\mathrm{H}_{10} = \frac{3.77 \text{ g}}{58.12 \text{ g/mol}} \approx 0.0649 \text{ mol}\]

Use Stoichiometry to Find Moles of \(\mathrm{O}_{2}\)

According to the balanced chemical equation, 2 moles of \( \mathrm{C}_{4}\mathrm{H}_{10} \) react with 13 moles of \( \mathrm{O}_{2} \). Using stoichiometry, we find the moles of \( \mathrm{O}_{2} \) needed:\[\text{Moles of } \mathrm{O}_{2} = 0.0649 \text{ mol } \mathrm{C}_4\mathrm{H}_{10} \times \frac{13 \text{ mol } \mathrm{O}_{2}}{2 \text{ mol } \mathrm{C}_4\mathrm{H}_{10}} \approx 0.4219 \text{ mol } \mathrm{O}_{2}\]

Convert Moles of \(\mathrm{O}_{2}\) to Liters at STP

To convert moles of \( \mathrm{O}_{2} \) to volume at STP (Standard Temperature and Pressure), use the conversion factor that 1 mole of any gas occupies 22.4 liters.\[\text{Liters of } \mathrm{O}_{2} = 0.4219 \text{ mol } \times 22.4 \text{ L/mol} \approx 9.45 \text{ L}\]

Key concepts

Moles of Butane

To understand the stoichiometry of a chemical reaction, we first need to find out how many moles of the substances involved are present. In this case, we're looking at butane, which is often used as a fuel in lighters. We start by calculating the moles of butane contained in the given mass.

The molar mass of butane (\( \mathrm{C}_4\mathrm{H}_{10} \)) is approximately 58.12 grams per mole. This value is critical since it allows us to convert a given mass into moles using the formula:\[\text{Moles of } \mathrm{C}_4\mathrm{H}_{10} = \frac{\text{mass of butane in g}}{\text{molar mass of butane in g/mol}}\]Suppose we have 3.77 grams of butane. By dividing this mass by the molar mass, we calculate the amount of butane in moles:\[\text{Moles of } \mathrm{C}_4\mathrm{H}_{10} = \frac{3.77 \text{ g}}{58.12 \text{ g/mol}} \approx 0.0649 \text{ mol}\]This computation helps in further stoichiometric calculations, linking the measurable macroscopic quantities, such as mass, to the number of particles, such as moles.

Balanced Chemical Equation

A balanced chemical equation is the heart of stoichiometry. It shows the relative proportions of reactants and products in a chemical reaction, represented in moles. Here, we are given the equation for the combustion of butane:\[2 \mathrm{C}_{4}\mathrm{H}_{10}(\mathrm{~g})+13 \mathrm{O}_{2}(\mathrm{~g}) \rightarrow 8 \mathrm{CO}_{2}(\mathrm{~g})+10 \mathrm{H}_{2} \mathrm{O}(\ell)\]This balanced equation tells us that 2 moles of butane react with 13 moles of oxygen to produce carbon dioxide and water. The coefficients (numbers in front of each molecule) represent the ratio of those molecules needed for the reaction to be balanced — meaning the same number of each type of atom appears on both sides of the equation.

Understanding this concept is key when calculating other quantities in the reaction. Know that if you have the moles of one substance (like butane), you can use these coefficients to find out how many moles of another substance (like oxygen) are involved via a simple ratio:\[\text{Moles of } \mathrm{O}_{2} = \text{Moles of } \mathrm{C}_4\mathrm{H}_{10} \times \frac{13 \text{ mol } \mathrm{O}_{2}}{2 \text{ mol } \mathrm{C}_4\mathrm{H}_{10}}\]This translation from moles of one substance to another ensures that everything adds up according to the law of conservation of mass.

Volume of Gas at STP

The concept of volume of gas at STP (Standard Temperature and Pressure) is crucial in chemistry, especially when linking the macroscopic properties of gases to their microscopic behaviors. At STP, the conditions are set to 0°C (273.15 K) temperature and 1 atm pressure. Under these conditions, one mole of an ideal gas occupies 22.4 liters. This relationship makes it straightforward to convert between moles and volume, using the conversion factor 1 mol = 22.4 L.

For instance, once you have calculated the moles of oxygen required for the combustion of butane as approximately 0.4219 moles, you can easily find out the volume of oxygen gas at STP using this simple multiplication:\[\text{Liters of } \mathrm{O}_{2} = \text{Moles of } \mathrm{O}_{2} \times 22.4 \text{ L/mol}\]Substituting the actual moles:\[0.4219 \text{ mol } \times 22.4 \text{ L/mol} \approx 9.45 \text{ L}\]This equation ensures a clear and straightforward understanding of how mole calculations in chemistry translate to real-world volumes at defined conditions, allowing practical and accurate predictions of how gases behave.