Problem 80 - Chemistry Matter and Change

Question

The reaction of chlorine gas with solid phosphorus \(\left(\mathrm{P}_{4}\right)\) produces solid phosphorus pentachloride. When 16.0 \(\mathrm{g}\) of chlorine reacts with 23.0 \(\mathrm{g}\) of \(\mathrm{P}_{4},\) which reactant is limiting? Which reactant is in excess?

Step-by-Step Solution

Answer

When 16.0 g of chlorine reacts with 23.0 g of \(P_4\), chlorine (\(Cl_2\)) is the limiting reactant, and phosphorus (\(P_4\)) is in excess.

Step 1: Write the balanced chemical equation

The reaction of chlorine gas with solid phosphorus produces solid phosphorus pentachloride. The balanced chemical equation for this reaction is: \[P_4 + 10Cl_2 \rightarrow 4PCl_5\]

Step 2: Convert the given masses of reactants to moles

We can use the molar mass of each reactant to convert the mass in grams to moles: Moles of \(P_4 = \frac{mass}{molar \, mass}\) Molar mass of \(P_4 = 4 \times 30.97 \, g/mol = 123.88 \, g/mol\) Moles of \(P_4 = \frac{23.0 \, g}{123.88 \, g/mol} = 0.1856 \, mol\) Moles of \(Cl_2 = \frac{mass}{molar \, mass}\) Molar mass of \(Cl_2 = 2 \times 35.45 \, g/mol = 70.90 \, g/mol\) Moles of \(Cl_2 = \frac{16.0 \, g}{70.90 \, g/mol} = 0.2256 \, mol\)

Step 3: Determine the limiting reactant and the reactant in excess

To compare the mole ratios, we will use the stoichiometry from the balanced chemical equation: \(1 mol \, P_4 : 10 mol \, Cl_2\) The given amounts of reactants are: \(0.1856 \, mol \, P_4 : 0.2256 \, mol \, Cl_2\) If we divide both mole amounts by their respective coefficients from the balanced equation, we can identify the limiting reactant: \(\frac{0.1856 \, mol \, P_4}{1} = 0.1856\) \(\frac{0.2256 \, mol \, Cl_2}{10} = 0.02256\) The smaller value corresponds to the limiting reactant. In this case, chlorine is the limiting reactant as 0.02256 < 0.1856. The reactant with the larger value is the reactant in excess, which is phosphorus (\(P_4\)).

Key Concepts

Balanced Chemical Equation

A balanced chemical equation is vital to accurately represent chemical reactions. Essentially, it describes what substances react and what products are formed, ensuring the law of conservation of mass is obeyed. This means the number of atoms for each element must be the same on both sides of the equation.
For example, in the reaction of phosphorus (\(P_4\)) with chlorine gas (\(Cl_2\)) to form phosphorus pentachloride (\(PCl_5\)), we start by writing down the unbalanced equation:\[P_4 + Cl_2 \rightarrow PCl_5\]Then, we balance it by adjusting the coefficients to account for the atoms:

• Each molecule of \(P_4\) contains 4 phosphorus atoms, requiring 4 molecules of \(PCl_5\); hence, 4 in front of \(PCl_5\).
• Each molecule of \(PCl_5\) needs 5 chlorine atoms. Since \(Cl_2\) provides 2 atoms per molecule, we need 10 molecules of \(Cl_2\) to supply the required 20 chlorine atoms.
• The balanced chemical equation becomes:\[P_4 + 10Cl_2 \rightarrow 4PCl_5\]

Balancing allows us to relate mole ratios, crucial for subsequent stoichiometric calculations.

Molar Mass

Molar mass is a key concept when calculating how substances react in exact amounts. It represents the mass of one mole (about 6.022 × 10²³ entities) of a given substance, expressed in grams per mole (g/mol). Knowing the molar mass enables us to convert between grams and moles, bridging the gap between laboratory measurements and theoretical contents of a reaction.
For phosphorus (\(P_4\)), first calculate the molar mass:

• One phosphorus atom has a molar mass of approximately 30.97 g/mol.
• Thus, \(P_4\) has a molar mass of 4 × 30.97 = 123.88 g/mol.

Similarly, for chlorine (\(Cl_2\)):

• A single chlorine atom weighs approximately 35.45 g/mol.
• \(Cl_2\) has a molar mass of 2 × 35.45 = 70.90 g/mol.

By converting grams to moles through division by the molar mass, we determine how many moles of each reactant are present, which is essential for further analysis in stoichiometry to find limiting reactants.

Stoichiometry

Stoichiometry is the method used to determine the quantitative relationships among reactants and products in a balanced chemical equation. It derives from the Greek words for "element" and "measure," and ensures that chemical reactions observe precise proportions.
This process involves several steps:

• Starting with the balanced chemical equation, identify the mole ratio between the reactants and products. From our example, the ratio is 1 mole of \(P_4\) to 10 moles of \(Cl_2\).
• Analyze the moles of each reactant to determine how they compare to this ratio. This is where you find which reactant is limiting since it will be completely consumed in the reaction, thus limiting the amount of product formed.
• In our case, given 0.1856 moles of \(P_4\) and 0.2256 moles of \(Cl_2\), the stoichiometry reveals chlorine is the limiting reactant. This is established by normalizing each reactant’s moles based on their balance coefficients (1 for \(P_4\) and 10 for \(Cl_2\)), showing \(Cl_2\) has the smallest ratio after applying its coefficient.

Recognizing the limiting reactant through stoichiometry helps in not just predicting the theoretical yield but also planning the quantities needed in chemical reactions.