Chapter 10: Problem 62
Calculate the mass in grams of hydrogen chloride produced when \(5.6 \mathrm{~L}\) of molecular hydrogen measured at STP react with an excess of molecular chlorine gas.
Short Answer
Expert verified
18.23 grams of HCl are produced.
Step by step solution
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Chemical Reactions
A chemical reaction involves the transformation of one or more substances into different substances. In this exercise, hydrogen gas (\( \text{H}_2 \)) and chlorine gas (\( \text{Cl}_2 \)) react to form hydrogen chloride (\( \text{HCl} \)). This is represented by the balanced chemical equation: \( \text{H}_2 + \text{Cl}_2 \rightarrow 2\text{HCl} \).
Balancing an equation ensures the same number of each type of atom on both sides. Here, one mole of \( \text{H}_2 \) reacts with one mole of \( \text{Cl}_2 \) to produce two moles of \( \text{HCl} \). This process follows the law of conservation of mass, indicating no atoms are lost or gained during the reaction. Understanding this helps identify how reactants convert to products.
Balancing an equation ensures the same number of each type of atom on both sides. Here, one mole of \( \text{H}_2 \) reacts with one mole of \( \text{Cl}_2 \) to produce two moles of \( \text{HCl} \). This process follows the law of conservation of mass, indicating no atoms are lost or gained during the reaction. Understanding this helps identify how reactants convert to products.
Mole Concept
The mole concept is a fundamental idea in chemistry that helps quantify amounts in a chemical reaction. A mole is a unit that represents \( 6.022 \times 10^{23} \) particles, such as atoms or molecules.
In this problem, we calculate the number of moles of hydrogen gas based on its volume at standard temperature and pressure (STP), where 1 mole of gas occupies \( 22.4 \text{ L} \). With \( 5.6 \text{ L} \) of \( \text{H}_2 \), we find it corresponds to \( 0.25 \) moles. This value helps us determine how much product, \( \text{HCl} \), can be formed. Moles allow us to connect mass and volume to the particles involved in the reaction.
In this problem, we calculate the number of moles of hydrogen gas based on its volume at standard temperature and pressure (STP), where 1 mole of gas occupies \( 22.4 \text{ L} \). With \( 5.6 \text{ L} \) of \( \text{H}_2 \), we find it corresponds to \( 0.25 \) moles. This value helps us determine how much product, \( \text{HCl} \), can be formed. Moles allow us to connect mass and volume to the particles involved in the reaction.
Ideal Gas Law
The ideal gas law is a useful tool to connect pressure, volume, and temperature of gases. The simple form used at STP (standard temperature and pressure) states that \( 1 \text{ mole} \) of an ideal gas occupies \( 22.4 \text{ L} \).
This principle helps us convert the given volume of hydrogen gas (\( 5.6 \text{ L} \)) to moles. It simplifies many stoichiometry problems involving gases, especially under standard conditions, by correlating the physical characteristics of gases with their chemical properties through the equation \( PV = nRT \). This allows for straightforward calculations when given volumes.
This principle helps us convert the given volume of hydrogen gas (\( 5.6 \text{ L} \)) to moles. It simplifies many stoichiometry problems involving gases, especially under standard conditions, by correlating the physical characteristics of gases with their chemical properties through the equation \( PV = nRT \). This allows for straightforward calculations when given volumes.
Molar Mass Calculation
Molar mass is the mass of one mole of a substance, often expressed in grams per mole. Calculating molar masses involves adding the atomic masses of all elements in a compound.
For \( \text{HCl} \), the molar mass is calculated by adding the atomic mass of hydrogen (\( 1.01 \text{ g/mol} \)) and chlorine (\( 35.45 \text{ g/mol} \)), resulting in \( 36.46 \text{ g/mol} \).
In the exercise, \( 0.5 \text{ moles} \) of \( \text{HCl} \) are produced. By multiplying by its molar mass, we find that the reaction yields approximately \( 18.23 \text{ g} \) of \( \text{HCl} \). Understanding molar mass connects the concept of moles with measurable quantities, such as mass.
For \( \text{HCl} \), the molar mass is calculated by adding the atomic mass of hydrogen (\( 1.01 \text{ g/mol} \)) and chlorine (\( 35.45 \text{ g/mol} \)), resulting in \( 36.46 \text{ g/mol} \).
In the exercise, \( 0.5 \text{ moles} \) of \( \text{HCl} \) are produced. By multiplying by its molar mass, we find that the reaction yields approximately \( 18.23 \text{ g} \) of \( \text{HCl} \). Understanding molar mass connects the concept of moles with measurable quantities, such as mass.
