Question

Elemental phosphorus reacts with chlorine gas according to the equation: $$ \mathrm{P}_{4}(s)+6 \mathrm{Cl}_{2}(g) \longrightarrow 4 \mathrm{PCl}_{3}(l) $$ A reaction mixture initially contains \(45.69 \mathrm{~g} \mathrm{P}_{4}\) and \(131.3 \mathrm{~g} \mathrm{Cl}_{2}\). Once the reaction has occurred as completely as possible, what mass (in g) of the excess reactant remains?

Short answer

7.45 g of P4 remains after the reaction.

Step-by-step solution

Calculate the molar mass of P4 and Cl2

To find the moles of P4 and Cl2 we need their molar masses. The molar mass of P4 is 4 times the atomic mass of phosphorus (30.974 g/mol), which equals 123.896 g/mol. The molar mass of Cl2 is 2 times the atomic mass of chlorine (35.453 g/mol), which equals 70.906 g/mol.

Find the moles of P4 and Cl2 present initially

Divide the masses of P4 and Cl2 by their respective molar masses to find the number of moles. Moles of P4 = 45.69 g / 123.896 g/mol = 0.369 moles. Moles of Cl2 = 131.3 g / 70.906 g/mol = 1.851 moles.

Determine the limiting reactant

Use the stoichiometry of the balanced equation P4 + 6Cl2 -> 4PCl3 to find the limiting reactant. The reaction requires 6 moles of Cl2 for every mole of P4. Calculate the required moles of Cl2 for the available P4: Required Cl2 = 0.369 moles P4 * (6 moles Cl2 / 1 mole P4) = 2.214 moles Cl2. Since we only have 1.851 moles of Cl2, Cl2 is the limiting reactant.

Calculate the mass of excess P4

Determine the moles of Cl2 that react completely with the available P4. Since Cl2 is limiting, use its mole quantity: Moles of P4 consumed = 1.851 moles Cl2 * (1 mole P4 / 6 moles Cl2) = 0.3085 moles P4. Calculate the mass of P4 consumed: Mass of P4 consumed = 0.3085 moles * 123.896 g/mol = 38.24 g. Subtract this from the initial mass to find the excess: Excess P4 = 45.69 g - 38.24 g = 7.45 g.

Report the mass of the excess reactant

Since Cl2 is the limiting reactant, P4 is in excess. The mass of P4 remaining after the reaction is 7.45 g.

Key concepts

Stoichiometry

Stoichiometry is the branch of chemistry that deals with the quantitative relationships between reactants and products in a chemical reaction. It is based on the conservation of mass and the law of definite proportions, which state that in a chemical reaction, matter is neither created nor destroyed, and elements combine in fixed ratios by mass.

When solving stoichiometry problems, one must first understand the balanced chemical equation, which provides the ratio in which reactants combine and products form. From there, mole-to-mole conversions help us predict how much of each reactant is needed and how much of each product will be formed. For instance, in the equation provided, one mole of phosphorus tetramer (), reacts with six moles of diatomic chlorine gas () to produce four moles of phosphorus trichloride ().

Stoichiometry is crucial in determining the limiting reactant, which is the substance that is completely consumed in the reaction and therefore determines the maximum amount of product that can form. It involves using mole ratios from the balanced equation to compare the moles of each reactant that would be needed versus the amount actually present. In the given exercise, the ratio of to needed is 1:6, hence by comparing the moles available, we identify which reactant limits the reaction.

Mole-to-Mass Calculations

Mole-to-mass calculations are a fundamental part of solving stoichiometry problems. They allow you to convert between the mass of a substance and the number of moles of that substance. Since chemical reactions are based on moles, being able to convert between mass and moles is essential to quantify the reactants and products involved.

To perform mole-to-mass calculations, you use the molar mass of a substance, which is the mass of one mole of that substance. The molar mass can be found by summing the atomic masses of elements in the compound, as found on the periodic table. In the example problem, the molar mass of is 123.896 g/mol and for , it is 70.906 g/mol. By dividing the given mass of each reactant by its molar mass, we can calculate the number of moles present before the reaction occurs.

This step is vital for understanding how much of each reactant will participate in the reaction. Using the stoichiometric coefficients from the balanced equation, we can then predict the amounts of products formed and the amounts of excess reactants remaining after the reaction completes.

Chemical Reaction Equations

Chemical reaction equations are a concise way of representing chemical reactions. These equations list the reactants on the left, typically separated by plus signs, and the products on the right, again separated by plus signs, with an arrow pointing from reactants to products to indicate the direction of the reaction.In a balanced chemical equation, each element has the same number of atoms on both sides of the equation, satisfying the law of conservation of mass. For example, the balanced equation in the exercise is which shows that one molecule reacts with six molecules to yield four molecules. Balancing chemical equations is crucial for accurately conducting mole-to-mass calculations and determining the limiting reactant in stoichiometric problems.

Understanding chemical equations allows us to visualize the reaction at the molecular level and provides the information needed to carry out quantitative analysis of the reaction. This information is crucial for scientists and engineers in various fields, including pharmacology, materials science, and environmental chemistry, where precise chemical formulations are essential.