Question

Elemental phosphorus reacts with chlorine gas according to the equation P4(s) + 6Cl2(g) → 4PCl3(l). A reaction mixture initially contains 91.42 g of P4 and 261.9 g of Cl2. What is the limiting reactant in this reaction?

1) P4
2) Cl2
3) PCl3
4) Cannot be determined

Answer 1

To find the limiting reactant, the masses of P4 and Cl2 are converted to moles and compared based on the stoichiometry of the balanced reaction. Cl2 is the limiting reactant because it has fewer moles than required to fully react with the available P4.

Step-by-step explanation:

To determine the limiting reactant in the chemical reaction P4(s) + 6Cl2(g) → 4PCl3(l), we need to compare the mole ratio of the reactants to the coefficients in the balanced equation. First, we need to convert the given masses of the reactants to moles.

For P4: The molar mass of P4 is approximately 123.89 g/mol (30.97 g/mol × 4). So, 91.42 g P4 ÷ 123.89 g/mol = 0.738 moles of P4.

For Cl2: The molar mass of Cl2 is approximately 70.90 g/mol (35.45 g/mol × 2). So, 261.9 g Cl2 ÷ 70.90 g/mol = 3.693 moles of Cl2.

According to the balanced equation, 1 mole of P4 reacts with 6 moles of Cl2. Based on the given amounts, 0.738 moles of P4 would require 4.428 moles of Cl2 (0.738 moles P4 × 6). However, there are only 3.693 moles of Cl2 available, which is less than the required 4.428 moles. Thus, Cl2 is the limiting reactant, since it will be completely consumed first.