Chapter 3: Problem 89
Consider the reaction $$ \mathrm{MnO}_{2}+4 \mathrm{HCl} \longrightarrow \mathrm{MnCl}_{2}+\mathrm{Cl}_{2}+2 \mathrm{H}_{2} \mathrm{O} $$ If \(0.86 \mathrm{~mol}\) of \(\mathrm{MnO}_{2}\) and \(48.2 \mathrm{~g}\) of \(\mathrm{HCl}\) react, which reactant will be used up first? How many grams of \(\mathrm{Cl}_{2}\) will be produced?
Short Answer
Expert verified
HCl is the limiting reactant; 46.86 g of Cl₂ will be 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.
Limiting Reactant
In chemical reactions, not all reactants are consumed equally. The limiting reactant is the substance that limits the extent of the reaction and determines how much product can be formed. It is the reactant that gets used up first. To find it, you need to establish the mole ratio from the balanced equation and compare the moles available for each reactant.
Once you find the moles of each reactant, you use the stoichiometric coefficients to understand which reactant gives you the least amount of product according to the balanced equation. The one that produces the least amount of product is the limiting reactant.
In our exercise, we found that HCl is the limiting reactant because fewer moles of HCl are available than required to react with the given moles of MnO extsubscript{2}. Identifying the limiting reactant is crucial as it directly impacts the amount of product formed in a reaction.
Once you find the moles of each reactant, you use the stoichiometric coefficients to understand which reactant gives you the least amount of product according to the balanced equation. The one that produces the least amount of product is the limiting reactant.
In our exercise, we found that HCl is the limiting reactant because fewer moles of HCl are available than required to react with the given moles of MnO extsubscript{2}. Identifying the limiting reactant is crucial as it directly impacts the amount of product formed in a reaction.
Mole Calculations
Mole calculations are a fundamental part of stoichiometry and involve determining the quantities of reactants and products in their moles. Moles help in expressing the amounts of chemicals involved in the reaction as they provide a bridge between the mass of a substance and the number of particles it contains.
To convert between mass and moles, we use the relation: \[ \text{Moles} = \frac{\text{Mass (g)}}{\text{Molar Mass (g/mol)}} \] For instance, in the given exercise, the molar mass of HCl is calculated, which is used to convert the mass of HCl to moles. This calculation is essential for understanding how much of a chemical substance is available to participate in a reaction.
Moles also allow you to use the mole ratio from the balanced equation to determine how reactants interact and how much product is formed. These calculations form the backbone of stoichiometry.
To convert between mass and moles, we use the relation: \[ \text{Moles} = \frac{\text{Mass (g)}}{\text{Molar Mass (g/mol)}} \] For instance, in the given exercise, the molar mass of HCl is calculated, which is used to convert the mass of HCl to moles. This calculation is essential for understanding how much of a chemical substance is available to participate in a reaction.
Moles also allow you to use the mole ratio from the balanced equation to determine how reactants interact and how much product is formed. These calculations form the backbone of stoichiometry.
Chemical Reactions
Chemical reactions involve the transformation of reactants into products through a systematic process that obeys the law of conservation of mass. This means that the same number of atoms of each element must be present on both sides of the chemical equation.
The balanced chemical equation that was given is: \[ \text{MnO}_{2} + 4 \text{HCl} \rightarrow \text{MnCl}_{2} + \text{Cl}_{2} + 2 \text{H}_{2} \text{O} \] It shows the stoichiometric relationships between reactants and products. Understanding these relationships is key to solving stoichiometry problems as it tells us how many moles of the products are formed from given moles of reactants.
This equation allows us to see that for every mole of MnO extsubscript{2} that reacts, four moles of HCl are necessary, producing one mole of MnCl extsubscript{2}, one mole of Cl extsubscript{2}, and two moles of H extsubscript{2}O. This demonstrates the stoichiometry that governs the proportions in which chemicals react and products form.
The balanced chemical equation that was given is: \[ \text{MnO}_{2} + 4 \text{HCl} \rightarrow \text{MnCl}_{2} + \text{Cl}_{2} + 2 \text{H}_{2} \text{O} \] It shows the stoichiometric relationships between reactants and products. Understanding these relationships is key to solving stoichiometry problems as it tells us how many moles of the products are formed from given moles of reactants.
This equation allows us to see that for every mole of MnO extsubscript{2} that reacts, four moles of HCl are necessary, producing one mole of MnCl extsubscript{2}, one mole of Cl extsubscript{2}, and two moles of H extsubscript{2}O. This demonstrates the stoichiometry that governs the proportions in which chemicals react and products form.
Mass-to-Mole Conversion
Mass-to-mole conversion is the process of converting the mass of a substance to moles using its molar mass. This conversion is a primary step in stoichiometry because it allows chemists to relate mass, a more practical measurement in the laboratory, to moles, which are more useful for theoretical calculations.
The formula used for mass-to-mole conversion is: \[ \text{Moles} = \frac{\text{Mass}}{\text{Molar Mass}} \] In the provided exercise, this conversion was applied to calculate the moles of HCl from its given mass. Using the molar mass of HCl, we found that 48.2 g of HCl converts to approximately 1.32 mol.
This conversion provided the groundwork for further calculations to determine the limiting reactant and the amount of product formed, ensuring that the stoichiometric relationships were preserved and accurately reflected in the reaction.
The formula used for mass-to-mole conversion is: \[ \text{Moles} = \frac{\text{Mass}}{\text{Molar Mass}} \] In the provided exercise, this conversion was applied to calculate the moles of HCl from its given mass. Using the molar mass of HCl, we found that 48.2 g of HCl converts to approximately 1.32 mol.
This conversion provided the groundwork for further calculations to determine the limiting reactant and the amount of product formed, ensuring that the stoichiometric relationships were preserved and accurately reflected in the reaction.
