Question

Phosphorous trichloride, a precursor for the manufacture of certain pesticides and herbicides, can be produced by the reaction of phosphorous and chlorine:

P₄ + 6Cl₂ => 4PCl₃

(a) When 20.5 g of P₄ and 20.5 g of Cl₂ are reacted, how many moles of PCl₃ can be produced?

Answer 1

When 20.5 g of P4 and 20.5 g of Cl2 are reacted, 0.66 moles of PCl3 can be produced, based on stoichiometry and limiting reagent calculations.

Step-by-step explanation:

The question deals with the stoichiometry of the reaction between phosphorus (P4) and chlorine (Cl2) to produce phosphorous trichloride (PCl3). To determine how many moles of PCl3 can be produced when 20.5 g of P4 are reacted with 20.5 g of Cl2, we first need to calculate the number of moles of each reactant using their respective molar masses.

The balanced chemical equation for the reaction is:

P4 + 6Cl2 → 4PCl3

Molar mass of P4 = 4 x 30.97 g/mol = 123.88 g/mol
Molar mass of Cl2 = 2 x 35.45 g/mol = 70.90 g/mol

Moles of P4 = 20.5 g / 123.88 g/mol = 0.165 moles
Moles of Cl2 = 20.5 g / 70.90 g/mol = 0.289 moles

Since phosphorus is the limiting reagent, the maximum moles of PCl3 produced will be dictated by the moles of P4 available. From the balanced equation, 1 mole of P4 produces 4 moles of PCl3, thus:

Moles of PCl3 = 0.165 moles P4 x (4 moles PCl3 / 1 mole P4) = 0.66 moles PCl3