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

After the reaction, 7.467 g of P4 remains as excess reactant.

Step-by-step solution

Calculate the Molar Mass of the Reactants

Determine the molecular weights of the reactants for Phosphorus (P4): Molar mass of P = 30.97 g/mol, so for P4 it's 4 * 30.97 g/mol = 123.88 g/mol. for Chlorine (Cl2): Molar mass of Cl = 35.45 g/mol, so for Cl2 it's 2 * 35.45 g/mol = 70.90 g/mol.

Determine the Moles of Each Reactant

Use the molar mass to convert the mass of each reactant to moles. For P4, 45.69 g P4 * (1 mol / 123.88 g) = 0.369 moles P4. For Cl2, 131.3 g Cl2 * (1 mol / 70.90 g) = 1.852 moles Cl2.

Identify the Limiting Reactant

The stoichiometry of the reaction shows that 1 mole of P4 reacts with 6 moles of Cl2. Thus, we need 0.369 moles P4 * 6 = 2.214 moles of Cl2 to fully react with the available P4. Since we only have 1.852 moles of Cl2, Cl2 is the limiting reactant.

Calculate the Amount of Excess Reactant

Since Cl2 is the limiting reactant and will be completely consumed, we need to find out how much excess P4 remains. The molar ratio allows us to find the theoretical amount of P4 needed, which is 1.852 moles Cl2 / 6 = 0.3087 moles P4. We initially had 0.369 moles P4, hence excess P4 = 0.369 - 0.3087 = 0.0603 moles P4.

Convert the Moles of Excess P4 to Mass

To find the mass of excess P4, we convert moles back to grams using the molar mass of P4. Excess mass of P4 = 0.0603 moles P4 * 123.88 g/mol = 7.467 g P4.

Key concepts

Limiting Reactant

Understanding the concept of a limiting reactant is crucial when studying chemical reactions. During a chemical reaction, reactants are converted into products, and the limiting reactant is the one that is completely consumed first, which prevents further reaction from occurring.

In every reaction, it's vital to identify the limiting reactant because it dictates how much product can be formed. To determine the limiting reactant, we compare the mole ratio of the reactants used in the reaction with the mole ratio of the reactants provided. If we take our sample equation, \(\mathrm{P}_{4}(s) + 6 \mathrm{Cl}_{2}(g) \longrightarrow 4 \mathrm{PCl}_{3}(l)\), the stoichiometry shows a 1:6 mole ratio between \(\mathrm{P}_{4}\) and \(\mathrm{Cl}_{2}\).

By calculating the moles of each reactant, and comparing those to the stoichiometric needs of the reaction, we identify the reactant which will run out first – the limiting reactant. This leaves the other reactant partially consumed – as the excess reactant, which will be left over after the reaction.

Molar Mass Calculation

The molar mass is a property that tells us how much one mole of a given substance weighs in grams. The molar mass is essential to convert between the mass of a substance and the number of moles. Calculating molar mass is straightforward when we know the atomic masses of the elements involved, which are found on the periodic table.

For instance, the molar mass of \(\mathrm{P}_{4}\) is calculated by multiplying the atomic mass of phosphorus (approximately 30.97 g/mol) by four, since there are four atoms of phosphorus in each molecule of \(\mathrm{P}_{4}\). Hence the molar mass of \(\mathrm{P}_{4}\) is 123.88 g/mol. Similarly, for \(\mathrm{Cl}_{2}\), we multiply the atomic mass of chlorine (approximately 35.45 g/mol) by two, resulting in a molar mass of 70.90 g/mol for \(\mathrm{Cl}_{2}\).

Knowing the molar masses allows the conversion of masses in grams to moles, which is a critical step in stoichiometric calculations.

Chemical Reaction Equation

A chemical reaction equation provides a symbolic representation of a chemical reaction. It lists the reactants and products, showing how the former are transformed into the latter. In our example, the equation \(\mathrm{P}_{4}(s) + 6 \mathrm{Cl}_{2}(g) \longrightarrow 4 \mathrm{PCl}_{3}(l)\) illustrates that one molecule of tetraphosphorus reacts with six molecules of chlorine gas to produce four molecules of phosphorus trichloride.

The stoichiometry of the reaction, given by the coefficients in the equation, tells us the exact proportions of reactants that combine and the amount of product that will form. Balancing this equation is critical because it ensures the conservation of mass and atoms, complying with the law of conservation of mass.

Mole-to-Mass Conversion

Mole-to-mass conversion is the process of converting the number of moles of a substance to the corresponding mass in grams using the molar mass of the substance. This step is fundamental in stoichiometry to relate quantitative information about the substances involved in a chemical reaction.

In the case of our reaction, once we have determined the number of moles of the excess reactant (\(\mathrm{P}_{4}\) in this case), we can easily find the remaining mass. We use the formula: \[\text{Mass} = \text{Moles} \times \text{Molar Mass}\]. Applying this to the excess \(\mathrm{P}_{4}\), with 0.0603 moles and a molar mass of 123.88 g/mol, the mass of excess \(\mathrm{P}_{4}\) is calculated to be approximately 7.467 g.

These conversions are fundamental in lab work and industry, where accurate amounts of substances are necessary to yield precise products, avoid waste, and ensure safety.