Thermal Decomposition: Gas Mixture Analysis

Hey guys, let's dive into a fascinating chemistry problem! We're going to break down (pun intended!) the thermal decomposition of an inorganic substance and analyze the resulting gas mixture. This kind of problem is super common in chemistry, and understanding how to approach it is key. So, grab your notebooks, and let's get started. We'll be looking at a scenario where 19.2 grams of an inorganic substance undergoes thermal decomposition, producing 13.44 liters of a gas mixture at standard conditions (that's what 'n.u.' means). We're also given that the relative density of this gas mixture with respect to hydrogen is 13. Finally, we know that when this mixture is passed over solid potassium hydroxide (KOH), the volume of the gas decreases by a factor of 1.5. Our goal? To figure out which gas is absorbed by the KOH.

Understanding Thermal Decomposition

Alright, first things first: thermal decomposition. This is a chemical reaction where a compound breaks down into simpler substances due to heat. Think of it like baking a cake – you apply heat (in the oven), and the cake's ingredients (flour, eggs, etc.) transform into a delicious whole. In our case, an inorganic substance is breaking down. Inorganic substances are compounds that typically don't contain carbon-hydrogen bonds, unlike the organic ingredients in our cake example. Common examples of inorganic substances include salts, oxides, and hydroxides. The products of thermal decomposition can be anything from gases to other solids, depending on the initial compound. The beauty of this is that it provides a very easy to understand and reliable way to analyze the substances. Keep in mind that everything depends on the context of the reaction.

Now, about those standard conditions ('n.u.'). Standard conditions (often denoted as STP, which stands for Standard Temperature and Pressure) means that the temperature is 0 degrees Celsius (273.15 Kelvin) and the pressure is 1 atmosphere (atm). At STP, one mole of any ideal gas occupies a volume of approximately 22.4 liters. This is a crucial piece of information, as it allows us to convert the volume of the gas mixture into moles, which we'll need later for our calculations. Having a strong grasp of these core concepts is essential for succeeding at this problem.

Calculating the Molar Mass of the Gas Mixture

Next, we have the relative density of the gas mixture with respect to hydrogen, which is 13. Relative density is simply the ratio of the density of a substance to the density of a reference substance (in this case, hydrogen). It's a dimensionless number. The relative density with respect to hydrogen can also be understood as the ratio of the molar mass of the gas mixture to the molar mass of hydrogen (which is 2 g/mol, since hydrogen gas exists as H2). So, to calculate the molar mass of the gas mixture, we do the following:

Molar mass of gas mixture = Relative density * Molar mass of hydrogen Molar mass of gas mixture = 13 * 2 g/mol = 26 g/mol

This tells us that, on average, a mole of the gas mixture has a mass of 26 grams. This will be very useful in finding the exact composition of the gas mixture later on. Remember that this is a weighted average, meaning that the molar mass reflects the relative amounts of each gas present. It's like finding the average weight of a group of people: if there are more heavier people, the average weight will be higher.

Determining the Volume of Gases Absorbed by KOH

Now, let's talk about potassium hydroxide (KOH). KOH is a strong base. It is a solid and it is highly soluble in water. It is excellent at absorbing acidic gases. The key here is to identify what gases are likely to react with a base like KOH. In general, acidic gases, such as carbon dioxide (CO2) and sulfur dioxide (SO2), will react with KOH and be absorbed, while inert gases, like nitrogen (N2) or oxygen (O2), will not. The problem states that the volume of the gas mixture decreased by a factor of 1.5 after passing over KOH. This means that a portion of the gas mixture was absorbed. To find out the volume of gas that was absorbed, we do the following:

Volume of absorbed gas = Initial volume of gas mixture - Final volume of gas mixture

Since the volume decreased by a factor of 1.5, the final volume of gas is:

Final volume = Initial volume / 1.5 Final volume = 13.44 L / 1.5 = 8.96 L

Therefore, the volume of gas absorbed by KOH is:

Volume of absorbed gas = 13.44 L - 8.96 L = 4.48 L

So, 4.48 liters of gas were absorbed by the KOH. Now we can start to figure out exactly what gases were in that absorbed volume. Keep in mind that we want to identify a specific compound.

Identifying the Absorbed Gas: The Solution

Since KOH absorbs acidic gases, we can hypothesize that the absorbed gas must have acidic properties. A gas like carbon dioxide (CO2) is a classic example of an acidic gas. Carbon dioxide reacts with a base like KOH to form potassium carbonate (K2CO3) and water. So, with that in mind, let's calculate the moles of the absorbed gas. At STP, 1 mole of any gas occupies 22.4 L. Therefore, the number of moles of absorbed gas can be calculated as follows:

Moles of absorbed gas = Volume of absorbed gas / Molar volume at STP Moles of absorbed gas = 4.48 L / 22.4 L/mol = 0.2 mol

So, 0.2 moles of gas were absorbed. Given our knowledge of the chemistry of KOH, the most likely gas absorbed is carbon dioxide (CO2). Carbon dioxide is a product of the decomposition of carbonates and bicarbonates, which are common inorganic compounds. Therefore, the gas absorbed by the KOH is CO2.

Putting it All Together

In conclusion, we've broken down the thermal decomposition problem step-by-step. We started with the basic concepts of thermal decomposition and standard conditions. We calculated the molar mass of the gas mixture using relative density. Then, we used the change in gas volume to determine the amount of gas absorbed by KOH. Finally, we used the behavior of KOH and our chemical knowledge to deduce that the absorbed gas was most likely carbon dioxide (CO2). This is a pretty common type of problem in introductory chemistry and highlights the importance of understanding chemical reactions, molar masses, and the properties of different compounds. Keep practicing these types of problems, and you'll become a pro in no time! Remember to always think through each step and consider the chemical properties of the substances involved. Good luck and happy studying!