Problem 101 - A Complete Resource Book in Chemistry for JEE Main

Question

'X' litres of carbon monoxide is present at STP. It is completely oxidized to \(\mathrm{CO}_{2}\). The volume of \(\mathrm{CO}_{2}\) formed is \(11.207\) litres at STP. What is the value of ' \(\mathrm{X}\) ' in litres? (a) \(32.2\) (b) \(21.2\) (c) \(10.2\) (d) \(11.2\)

Step-by-Step Solution

Answer

Option (d) 11.2 litres is closest to the calculated volume of X, which is 11.207 litres.

Step 1: Understanding the Reaction

The reaction involved is the oxidation of carbon monoxide to carbon dioxide. The balanced chemical equation is \[ \text{2CO} + \text{O}_2 \rightarrow \text{2CO}_2 \]. From this equation, we see that 2 volumes of CO produce 2 volumes of CO_2 at STP.

Step 2: Relation at STP

According to the balanced equation, the volume of CO is equal to the volume of CO_2 produced because 2 litres of CO produces 2 litres of CO_2. This implies that the volume of CO is equal to the volume of CO_2 under Standard Temperature and Pressure conditions.

Step 3: Solving for X

Since the equation shows that the volume of CO equals the volume of CO_2, and we know the volume of CO_2 is given as 11.207 litres, it means the volume of CO, denoted as X, is also 11.207 litres. Thus, \( X = 11.207 \) litres.

Step 4: Choosing the Closest Option

The closest option to 11.207 litres from the provided choices is (d) 11.2 litres.

Key Concepts

Chemical reactions

Chemical reactions involve the transformation of substances through the making and breaking of chemical bonds. In the context of gases, this often means observing how reactants are converted into different gaseous products. A balanced chemical equation provides the roadmap for understanding the proportions of reactants and products. For our particular reaction, carbon monoxide (CO) is oxidized to form carbon dioxide (CO_2). This process involves the reaction of CO with oxygen (O_2). The balanced equation, \[ \text{2CO} + \text{O}_2 \rightarrow \text{2CO}_2 \], indicates that two molecules of carbon monoxide react with one molecule of oxygen to form two molecules of carbon dioxide. This balanced equation is crucial as it preserves the law of conservation of mass, stating that mass is neither created nor destroyed in a chemical reaction. Knowing the proportions in the balanced equation allows us to predict the outcome of the reaction under different conditions.

Gas laws

Gas laws describe how gases behave under various conditions of pressure, volume, and temperature. The most fundamental of these is the ideal gas law, which is usually expressed as \( PV = nRT \). This equation relates pressure (P), volume (V), the number of moles of gas (n), the ideal gas constant (R), and temperature (T). A key aspect of stoichiometry in gas reactions is understanding how these variables interact when a reaction occurs.In the specific reaction where carbon monoxide is oxidized to form carbon dioxide, it is helpful to know the volumes of gases involved. Since we are dealing with gases, we often use the concept of mole-to-mole ratios based on the balanced chemical equation to determine the resulting volumes. The chemical reaction follows Avogadro's law, which states that equal volumes of gases at the same temperature and pressure contain an equal number of molecules. Hence, if the volume of CO_2 produced at the end of the reaction is known, and it equals the volume of CO used, then we can infer the initial volume of CO from the volume of CO_2 formed.

Standard Temperature and Pressure

Standard Temperature and Pressure (STP) are conditions often used as a reference point in gas calculations. STP is defined as a temperature of 0°C (273.15 K) and a pressure of 1 atm. These conditions simplify calculations because, at STP, one mole of an ideal gas occupies 22.414 litres. This relationship helps in determining volumes of gases in reactions that follow ideal gas behavior. In our exercise, knowing that all processes are taking place at STP simplifies the stoichiometric calculations. For example, since the volume of carbon monoxide is equal to the resultant volume of carbon dioxide under STP, and given that the volume of CO_2 is 11.207 litres, it follows directly that the volume of CO must also have been 11.207 litres. Using these STP values allows for straightforward conversion between amounts of gas in moles and gas volumes, easing the determination of quantities involved in chemical reactions.