Question

Consider the following reaction for the combustion of octane, \(\mathrm{C}_{8} \mathrm{H}_{18}\) : $$ 2 \mathrm{C}_{8} \mathrm{H}_{18}(l)+25 \mathrm{O}_{2}(g) \rightarrow 16 \mathrm{CO}_{2}(g)+18 \mathrm{H}_{2} \mathrm{O}(l) $$ What volume of oxygen gas at STP would be needed for the complete combustion of \(10.0 \mathrm{~g}\) of octane?

Short answer

24.50 liters of oxygen gas at STP are needed for the complete combustion of 10.0 g of octane.

Step-by-step solution

Calculate moles of octane

To convert the mass of octane (10.0 g) into moles, we need to know its molar mass. The molar mass of octane (C8H18) can be calculated as: Molar mass = (8 × 12.01 g/mol for C) + (18 × 1.01 g/mol for H) = 114.23 g/mol Now we can find out the moles of octane: Moles of octane = Mass / Molar mass = \( \frac{10.0 \ g}{114.23 \ g/mol} \) = 0.0875 mol Step 2: Determine moles of oxygen needed

Determine moles of oxygen needed

From the balanced chemical equation, 2 moles of octane react with 25 moles of oxygen. Therefore, we can use the stoichiometry to find the moles of oxygen needed for the given moles of octane: Moles of oxygen = moles of octane × \( \frac{25 \ moles\ of\ O_2}{2 \ moles\ of\ C_8H_{18}} \) = 0.0875 mol × \( \frac{25}{2} \) = 1.094 mol Step 3: Calculate the volume of oxygen gas at STP

Calculate the volume of oxygen gas at STP

At STP (standard temperature and pressure: 0 °C and 1 atm), 1 mole of any ideal gas occupies a volume of 22.4 liters. We can use this fact to convert the moles of oxygen obtained in step 2 to the volume: Volume of oxygen = moles of oxygen × 22.4 L/mol = 1.094 mol × 22.4 L/mol = 24.50 L Therefore, 24.50 liters of oxygen gas at STP are needed for the complete combustion of 10.0 g of octane.

Key concepts

Understanding Stoichiometry

Stoichiometry is the branch of chemistry that deals with the quantitative relationships between reactants and products in a chemical reaction. Think of it as the recipe for chemistry, where one needs to measure the exact amount of ingredients (reactants) to get the desired output (products). In the combustion of octane, we see that 2 moles of octane react with 25 moles of oxygen to produce carbon dioxide and water. Hence, stoichiometry helps us to predict how much oxygen is required to completely burn a given amount of octane. Through careful calculation, we can avoid the wastage of reactants and ensure that a reaction goes to completion.

Let's make it even simpler. If you were baking cookies and the recipe calls for 2 eggs for every cup of flour, what would you do if you only had half a cup of flour? You’d use just one egg. Stoichiometry uses this same principle but involves molecules and moles instead of eggs and flour.

Navigating Molar Mass Calculations

Knowing the molar mass of a substance is akin to understanding how much a currency is worth before you can buy anything. For octane, which is made up of carbon and hydrogen atoms, we calculate its molar mass by adding up the molar masses of all the atoms present.

We find the molar mass by taking the atomic mass of carbon (12.01 g/mol), multiplying it by the number of carbon atoms (8), and then doing the same for hydrogen (1.01 g/mol, multiplied by 18 hydrogen atoms). This gives us the molar mass of octane. Using this crucial piece of information, we then convert grams of octane into moles, since chemical reactions are based on moles rather than grams. This step is paramount in the journey of understanding how much of each substance is required or produced in a reaction.

Applying Gas Laws to Determine Volumes

Gas laws are the set of laws that describe the behavior of gases, and one essential gas law is related to the conditions of Standard Temperature and Pressure (STP). This simplification allows for the conclusion that any ideal gas will occupy 22.4 liters per mole at STP. Why does this matter for the combustion of octane?

When we've found out how many moles of oxygen we need, we apply the gas law to determine the volume that these moles of oxygen would occupy at STP. Like fitting air into a balloon, gas laws let us predict how much space the gas needs. If you’re blowing up a balloon in a cold room (at standard temperature), you’d expect it to have a set size. That's what STP is—a specific, predictable condition for measuring gases. By mastering gas laws, we're able to link the amount of substance directly to a volume we can measure.