Question

Classify the type of stoichiometry problem: How many liters of carbon dioxide gas are produced from the decomposition of \(25.0 \mathrm{~g} \mathrm{NaHCO}_{3} ?\)

Short answer

It is a mass-to-gas volume stoichiometry problem; 3.34 L of CO₂ is produced.

Step-by-step solution

Understand the Problem

First, identify the chemical reaction involved. Sodium bicarbonate (NaHCO₃) decomposes into sodium carbonate (Na₂CO₃), carbon dioxide (CO₂), and water (H₂O). The balanced chemical equation is: \[ 2 ext{NaHCO}_3 ightarrow ext{Na}_2 ext{CO}_3 + ext{CO}_2 + ext{H}_2 ext{O} \]

Identify the Type of Stoichiometry Problem

This problem involves converting mass to volume using stoichiometry, starting from a known mass of a reactant (NaHCO₃) and ending with the volume of a gaseous product (CO₂). This falls under the category of mass-to-gas volume stoichiometry.

Calculate Moles of NaHCO₃

Calculate the number of moles of NaHCO₃ using its molar mass. The molar mass of NaHCO₃ is approximately 84.01 g/mol. Use the formula: \[ \text{Moles of NaHCO}_3 = \frac{25.0 \, ext{g}}{84.01 \, ext{g/mol}} \approx 0.298 \text{ moles}\]

Determine Moles of CO₂ Produced

Using the balanced equation, determine the stoichiometry. From the balanced equation, 2 moles of NaHCO₃ produce 1 mole of CO₂: \[ \text{Moles of CO}_2 = \frac{0.298 \text{ moles of NaHCO}_3}{2} \approx 0.149 \text{ moles of CO}_2 \]

Calculate Volume of CO₂ Produced

Use the ideal gas law assumption that at standard temperature and pressure (STP), 1 mole of gas occupies 22.4 liters. Therefore, the volume of CO₂ is: \[ \text{Volume of CO}_2 = 0.149 \text{ moles} \times 22.4 \frac{L}{ ext{mole}} = 3.34 \text{ L} \]

Summarize the Solution

The problem has been classified as a mass-to-gas volume stoichiometry problem, and the volume of CO₂ gas produced is calculated based on the decomposition reaction, using stoichiometry to connect mass and volume.

Key concepts

Chemical Reactions and Decomposition

In the world of chemistry, reactions are all about breaking and forming bonds. A chemical reaction involves substances—called reactants—changing into other substances, known as products. In the reaction from our exercise, we're dealing with the decomposition of sodium bicarbonate (NaHCO₃). When sodium bicarbonate decomposes, it transforms into three different products: sodium carbonate (Na₂CO₃), carbon dioxide (CO₂), and water (H₂O). This is a classic example of a decomposition reaction, where a single compound breaks down into multiple smaller compounds. The balanced equation for this reaction is crucial because it provides the exact proportions of reactants and products needed. Understanding this balanced equation, \[2 \text{NaHCO}_3 \rightarrow \text{Na}_2\text{CO}_3 + \text{CO}_2 + \text{H}_2\text{O}\]means recognizing that 2 moles of NaHCO₃ produce 1 mole each of Na₂CO₃, CO₂, and H₂O. This mole relationship is the foundation of what we call stoichiometry in chemistry.

Mass-to-Volume Conversion in Stoichiometry

Stoichiometry is the calculation of relative quantities of reactants and products in chemical reactions. Our exercise involves a type of stoichiometry known as mass-to-volume conversion, specifically for gases. The goal is to start with a known mass of a reactant (in this case, 25 grams of NaHCO₃) and find out how much volume of a gaseous product (CO₂) is formed. Here's how it typically works:

• First, you find the moles of the reactant. This is done by dividing the mass of the compound by its molar mass. For NaHCO₃, the molar mass is approximately 84.01 g/mol.


• Next, using the balanced chemical equation, you find out how many moles of the product will form from the reactant. This is where stoichiometry plays a big role, using the mole-to-mole ratios from the equation.


• Finally, convert these moles to volume. For gases, this is often done using the ideal gas law relationship at standard temperature and pressure (STP), where 1 mole of any ideal gas occupies 22.4 liters.

By following these steps, you bridge the gap between mass (a tangible quantity) and volume (a measure of space the gas occupies), using mathematical conversions based on the properties of gases.

Understanding the Ideal Gas Law

The ideal gas law is a fundamental concept in chemistry and deals with the behavior of gases under various conditions. In our stoichiometry problem, we rely on this principle to convert moles of CO₂ gas into volume.The ideal gas law is often simplified to the condition of standard temperature and pressure (STP) for ease of calculation in stoichiometric problems. At STP (0 degrees Celsius and 1 atm pressure), 1 mole of an ideal gas occupies a volume of 22.4 liters.This simplification becomes particularly handy in our exercise. Once we found the moles of CO₂ formed from the decomposition reaction, we used the ratio from the ideal gas law to find out how many liters this corresponds to:\[\text{Volume of CO}_2 = 0.149 \text{ moles} \times 22.4 \frac{L}{\text{mole}} = 3.34 \text{ L}\]Remember:

• The ideal gas law is a theoretical model and best applies under low pressure and high temperature.


• While real gases aren't perfect, the law provides reliable estimates in many common situations involving gases.


• When you're working with problems like this, always check if conditions approximate STP, or adjust calculations according to the actual conditions given.