Question

When solid phosphorus \(\left(\mathrm{P}_{4}\right)\) reacts with oxygen gas, diphosphorus pentoxide is formed. Initially \(2.87 \mathrm{~mol}\) of phosphorus and \(3.86 \mathrm{~mol}\) of oxygen are combined. (Assume \(100 \%\) yield.) (a) Write a balanced equation for the reaction. (b) What is the limiting reactant? (c) How many moles of excess reactant remain after the reaction is complete?

Short answer

Question: Solid phosphorus (P4) reacts with oxygen gas to form diphosphorus pentoxide. Given 2.87 moles of P4 and 3.86 moles of O2, determine the limiting reactant and the moles of excess reactant left after the reaction is complete. Answer: The limiting reactant is O2 and there are 2.098 moles of excess reactant (P4) remaining after the reaction is complete.

Step-by-step solution

(a) Balanced equation for the reaction

To balance the equation for this reaction, we need to make sure there are equal numbers of atoms on both sides: P4 + 5O2 -> 2P2O5 Now the equation is balanced.

(b) Finding the limiting reactant

We have 2.87 mol of P4 and 3.86 mol of O2. From the balanced equation, we can see that the molar ratio between P4:O2 is 1:5. First, we need to find the amount of O2 that is required for the 2.87 mol of P4 to react completely: \(\text{mole ratio } = \frac{\text{mol of O2 required}}{\text{2.87 mol of P4}} = \frac{5 \text{ mol O2}}{1 \text{ mol P4}}\) Now, we solve for the moles of O2 required: \(\text{mol of O2 required} = \text{mole ratio} × \text{mol of P4}\) \(\text{mol of O2 required} = \frac{5\cdot2.87}{1} = 14.35 \text{ mol O2}\) Since there is 3.86 mol of O2 present but we require 14.35 mol of O2 for all the P4 to react, it's clear that O2 will be the limiting reactant. The limiting reactant is O2.

(c) Determining moles of excess reactant remaining

To find the number of moles of excess reactant (P4) remaining after the reaction, we'll use the limiting reactant (O2) to calculate the amount of P4 that was consumed during the reaction: \(\text{mole ratio (P4 consumed to O2)} = \frac{1 \text{ mol P4}}{5 \text{ mol O2}}\) \(\text{mol P4 consumed} = \text{mole ratio} × \text{mol of O2}\) \(\text{mol P4 consumed} = \frac{1\cdot3.86}{5} = 0.772 \text{ mol P4}\) Now, we can calculate the moles of excess reactant (P4) remaining by subtracting the consumed amount from the initial amount: \(\text{mol P4 remaining} = \text{initial mol P4} - \text{mol P4 consumed} = 2.87 \text{ mol } - 0.772 \text{ mol} = 2.098 \text{ mol P4}\) There are 2.098 moles of excess reactant (P4) remaining after the reaction is complete.