Chapter 7: Problem 56
Elemental phosphorus reacts with chlorine gas according to the equation: P4(s) + 6 Cl2(g)-4 PCl3(l) A reaction mixture initially contains 45.69 g P4 and 131.3 g Cl2. Once the reaction has occurred as completely as possible, what mass (in g) of the excess reactant remains?
Short Answer
Expert verified
Remaining mass of Cl2 is 32.52 g.
Step by step solution
01
Determine the Molar Masses
Calculate the molar mass of the reactants. For phosphorus (P4), the molar mass is 123.895 g/mol (30.974 g/mol for P times 4). For chlorine gas (Cl2), the molar mass is 70.906 g/mol (35.453 g/mol for Cl times 2).
02
Convert Masses to Moles
Use the molar masses to convert the given masses to moles. For P4: \[\begin{equation} \text{Moles P4} = \frac{45.69 \, \text{g}}{123.895 \, \text{g/mol}} \end{equation}\]. For Cl2: \[\begin{equation} \text{Moles Cl2} = \frac{131.3 \, \text{g}}{70.906 \, \text{g/mol}} \end{equation}\]
03
Determine the Limiting Reactant
Use the stoichiometry of the balanced equation to find which reactant is the limiting reactant. According to the equation, 1 mole of P4 reacts with 6 moles of Cl2. Compare the mole ratio of the reactants to find the limiting reactant.
04
Calculate the Moles of Excess Reactant
Subtract the moles of Cl2 that were actually used (based on the stoichiometric ratio with the limiting reactant) from the initial moles of Cl2 to find the moles of Cl2 remaining.
05
Convert Moles of Excess Reactant to Mass
Convert the remaining moles of Cl2 back to mass using its molar mass. Multiply the moles of the excess reactant by its molar mass to obtain the mass of excess Cl2 that remains after the reaction.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Stoichiometry
Stoichiometry is akin to the recipe for a chemical reaction; it equates the proportions of reactants and products during a chemical process. By using the coefficients from the balanced equation, we can determine the exact amount of reactants necessary to produce a certain amount of product.
For instance, in the equation for the reaction between elemental phosphorus and chlorine gas to form phosphorus trichloride, a stoichiometric coefficient tells us that for every molecule of P4, 6 molecules of Cl2 are required to react completely. Understanding this ratio is crucial, as it allows us to identify the limiting reactant — the reactant that will be consumed first, thus determining the maximum amount of product that can be formed. In this case, by comparing the molar amounts of P4 and Cl2 that we start with, we can figure out which will limit the reaction and what quantity of the other reactant will be in excess.
Stoichiometry is not just theoretical; it has very practical applications. For example, it can help industries optimize the use of reactants in large-scale productions to minimize waste and cost. Knowing how to perform stoichiometric calculations is essential for anyone studying chemistry—whether in academic settings or in the field.
For instance, in the equation for the reaction between elemental phosphorus and chlorine gas to form phosphorus trichloride, a stoichiometric coefficient tells us that for every molecule of P4, 6 molecules of Cl2 are required to react completely. Understanding this ratio is crucial, as it allows us to identify the limiting reactant — the reactant that will be consumed first, thus determining the maximum amount of product that can be formed. In this case, by comparing the molar amounts of P4 and Cl2 that we start with, we can figure out which will limit the reaction and what quantity of the other reactant will be in excess.
Stoichiometry is not just theoretical; it has very practical applications. For example, it can help industries optimize the use of reactants in large-scale productions to minimize waste and cost. Knowing how to perform stoichiometric calculations is essential for anyone studying chemistry—whether in academic settings or in the field.
Molar Mass Calculation
Molar mass is the weight of one mole of a substance, expressed in grams per mole (g/mol). It's a bridge between the atomic and macroscopic worlds. To find the molar mass of a compound, we sum up the masses of the individual atoms that make up the molecule.
In the given exercise, to calculate the molar mass of phosphorus (P4), we multiplied the atomic mass of phosphorus (30.974 g/mol) by 4, because there are four phosphorus atoms in the molecule. The process for chlorine (Cl2) is similar, multiplying the atomic mass of chlorine (35.453 g/mol) by 2. These calculations are fundamental, as they allow us to convert the amounts of reactants from grams to moles—a format necessary for comparing the quantities in terms of the stoichiometry of the reaction.
It's also important to remember that molar mass can vary slightly based on isotopic composition, which can differ between samples. However, for most educational purposes and calculations, we use the average atomic mass listed on the periodic table.
In the given exercise, to calculate the molar mass of phosphorus (P4), we multiplied the atomic mass of phosphorus (30.974 g/mol) by 4, because there are four phosphorus atoms in the molecule. The process for chlorine (Cl2) is similar, multiplying the atomic mass of chlorine (35.453 g/mol) by 2. These calculations are fundamental, as they allow us to convert the amounts of reactants from grams to moles—a format necessary for comparing the quantities in terms of the stoichiometry of the reaction.
It's also important to remember that molar mass can vary slightly based on isotopic composition, which can differ between samples. However, for most educational purposes and calculations, we use the average atomic mass listed on the periodic table.
Chemical Reaction Equation
A chemical reaction equation is more than just a sentence; it's a depiction of a chemical transformation. It provides a wealth of information such as the reactants, products, states of these substances (solid, liquid, gas), and the proportions in which they interact and form. It's a balance of mass and charge because in a chemical reaction, matter cannot be created or destroyed, and atoms must be conserved.
In our case with elemental phosphorus and chlorine gas, the reaction equation is P4(s) + 6 Cl2(g) → 4 PCl3(l). This balanced equation shows that four phosphorus atoms (as part of the P4 molecule) and twelve chlorine atoms (from six Cl2 molecules), undergo a chemical change to form four PCl3 molecules.
Understanding this equation thoroughly allows us not only to predict the products but also to calculate the quantities involved — this is where stoichiometry comes into play. When approaching a problem, verifying that the equation is balanced is always the first step before any stoichiometric computations can be performed. A good grasp of chemical reaction equations is essential for anyone delving into the study of chemistry or related fields.
In our case with elemental phosphorus and chlorine gas, the reaction equation is P4(s) + 6 Cl2(g) → 4 PCl3(l). This balanced equation shows that four phosphorus atoms (as part of the P4 molecule) and twelve chlorine atoms (from six Cl2 molecules), undergo a chemical change to form four PCl3 molecules.
Understanding this equation thoroughly allows us not only to predict the products but also to calculate the quantities involved — this is where stoichiometry comes into play. When approaching a problem, verifying that the equation is balanced is always the first step before any stoichiometric computations can be performed. A good grasp of chemical reaction equations is essential for anyone delving into the study of chemistry or related fields.