Question

What mass of carbon must burn to produce 4.56 \(\mathrm{L} \mathrm{CO}_{2}\) gas at STP? (Chapter 11) $$ \mathrm{C}(\mathrm{s})+\mathrm{O}_{2}(\mathrm{g}) \rightarrow \mathrm{CO}_{2}(\mathrm{g}) $$

Short answer

The mass of carbon required to produce 4.56 L of CO2 gas at STP can be calculated using the ideal gas law and mole concept. First, find the moles of CO2 produced: moles of CO2 = 4.56 L / 22.4 L/mol. Then, as the balanced equation shows a 1:1 ratio between carbon and CO2, the moles of carbon required are equal to the moles of CO2. Finally, calculate the mass of carbon: mass of C = moles of C × 12.01 g/mol. Thus, approximately 2.44 grams of carbon must burn to produce 4.56 L of CO2 gas at STP.

Step-by-step solution

Convert volume of CO2 to moles

At STP, one mole of any gas occupies 22.4 L of volume. To find the moles of CO2 produced, we will use the formula: moles of CO2 = volume of CO2 / volume occupied by one mole at STP moles of CO2 = 4.56 L / 22.4 L/mol

Calculate the moles of carbon required

According to the balanced equation, one mole of carbon burns to produce one mole of CO2. Therefore, the moles of carbon required will be equal to the moles of CO2 produced. moles of C = moles of CO2

Calculate the mass of carbon required

Now that we have the moles of carbon required, we can find the mass using the formula: mass = moles × molar mass The molar mass of carbon is 12.01 g/mol. So, mass of C = moles of C × 12.01 g/mol

Find the mass of carbon required to produce 4.56 L CO2

Now, simply plug the values from the previous steps into the formula: mass of C = (4.56 L / 22.4 L/mol) × 12.01 g/mol mass of C = 0.2036 mol × 12.01 g/mol mass of C ≈ 2.44 g So, approximately 2.44 grams of carbon must burn to produce 4.56 L of CO2 gas at STP.

Key concepts

Moles

In the world of chemistry, moles are a fundamental unit that makes it easier for scientists to quantify substances. A mole represents a standard number of particles, much like a dozen represents twelve. Specifically, one mole contains Avogadro's number of particles: approximately 6.022 x 10^23. This could be atoms, molecules, ions, etc.
Understanding moles helps us move between the atomic scale and the human scale. It helps chemists quantify reactions and make precise measurements. For gases, the mole concept is invaluable, especially at standard conditions.
Using moles, we can calculate how many atoms or molecules are involved in a reaction. It ties together both the macroscopic amounts we can measure and the microscopic amounts we cannot see. In chemical equations, using moles helps balance the number of atoms reacting and being formed, ensuring that mass remains conserved.

STP (Standard Temperature and Pressure)

Standard Temperature and Pressure, or STP, is a set of conditions used universally in scientific studies to allow for consistency and comparability in experiments. At STP, the temperature is 0°C (273.15 K) and the pressure is 1 atmosphere (atm). These conditions are defined because they are easy to replicate and provide a common ground for scientists to compare results.
At STP, gases take up a predictable volume, making calculations in stoichiometry straightforward. For instance, one mole of any gas will occupy 22.4 liters. This rule significantly simplifies the process of converting between the volume of gas and the amount of substance involved in chemical reactions.
Using STP ensures that researchers and students can communicate findings and understand problem sets without becoming bogged down by inconsistencies in measurement conditions. It creates a universal language for the description of gas behavior.

Chemical Equations

Chemical equations are like recipes for reactions; they tell us what substances react and what products are formed, as well as in what proportion. For example, in the reaction: \( \mathrm{C}(\mathrm{s})+\mathrm{O}_{2}(\mathrm{g}) \rightarrow \mathrm{CO}_{2}(\mathrm{g}) \), carbon reacts with oxygen gas to form carbon dioxide gas.
These equations must be balanced, meaning each type of atom must appear the same number of times on both sides of the equation. This is crucial because it adheres to the Law of Conservation of Mass, which asserts that matter cannot be created or destroyed.
Balancing chemical equations ensures that the quantity of reactants used equals the quantity of products made. It provides a clear framework for calculating how much of each substance is needed or produced in a given reaction, making tasks like predicting the mass of reactants or products required much easier.