Equal Masses of phosphorus and chlorine react according to the follow equation.
2P+3Cl2=2PCl3
Which reagent is in excess and what percentage of it will remain at the end of the reaction
To determine which reagent is in excess and what percentage of it will remain at the end of the reaction, we need to compare the stoichiometry of the two reactants.
The balanced equation is:
2P + 3Cl2 → 2PCl3
From the equation, we can see that for every 3 moles of chlorine (Cl2), we need 2 moles of phosphorus (P) to react completely.
Let's assume we have equal masses of phosphorus and chlorine, so we have the same number of moles for each.
To find out which reagent is in excess, we need to calculate the moles of each reagent.
Step 1: Convert the masses of phosphorus and chlorine to moles.
Assuming the mass of both phosphorus and chlorine is "x" grams, we can use their molar masses to find the number of moles:
Molar mass of phosphorus (P) = 31.0 g/mol
Molar mass of chlorine (Cl2) = 70.9 g/mol
Moles of phosphorus (P) = Mass of phosphorus (x g) / Molar mass of phosphorus (31.0 g/mol)
= x/31.0 mol
Moles of chlorine (Cl2) = Mass of chlorine (x g) / Molar mass of chlorine (70.9 g/mol)
= x/70.9 mol
Step 2: Compare the stoichiometric ratio of the two reactants.
From the balanced equation, we know that 2 moles of phosphorus react with 3 moles of chlorine.
So, the ratio of moles of phosphorus to moles of chlorine is 2:3.
Step 3: Determine the limiting reagent by comparing the ratios.
To determine which reagent is in excess, we compare the ratio of actual moles of each reagent to the stoichiometric ratio.
Actual ratio of moles of phosphorus to moles of chlorine = (x/31.0) / (x/70.9)
= 70.9 / 31.0
≈ 2.286
Since the actual ratio is slightly more than the stoichiometric ratio (2:3), phosphorus (P) is the limiting reagent, and chlorine (Cl2) is in excess.
Step 4: Calculate the moles of excess reagent and the percentage remaining.
Since phosphorus is the limiting reagent, we can calculate the moles of phosphorus trichloride (PCl3) formed by using the stoichiometric ratio.
From the equation, 2 moles of phosphorus react to form 2 moles of phosphorus trichloride (PCl3).
So, moles of PCl3 formed = 2 moles
Now, to find the moles of excess chlorine remaining, we can subtract the moles of phosphorus trichloride formed from the initial moles of chlorine.
Moles of chlorine remaining = Moles of chlorine initially - Moles of PCl3 formed
= (x/70.9) - 2
To calculate the percentage of chlorine remaining, we can use the mass of chlorine remaining and the total mass of chlorine initially:
Mass of chlorine remaining = Moles of chlorine remaining x Molar mass of chlorine (70.9 g/mol)
Percentage remaining = (Mass of chlorine remaining / Mass of chlorine initially) x 100
Plug in the value for moles of chlorine remaining from the previous step, and solve for the percentage.
This calculation will give us the percentage of chlorine remaining at the end of the reaction when phosphorus is the limiting reagent.
To determine which reagent is in excess and the percentage of it that will remain at the end of the reaction, you can use the concept of limiting reactants.
1. Start by converting the masses of both phosphorus (P) and chlorine (Cl2) to moles. This can be done by using the molar mass of each element:
Molar mass of P = 31 g/mol
Molar mass of Cl2 = 71 g/mol
2. Assuming you have an equal mass for both P and Cl2, let's say 1 gram each. Using the molar masses, we can convert these masses into moles:
Moles of P = mass of P / molar mass of P = 1 g / 31 g/mol
Moles of Cl2 = mass of Cl2 / molar mass of Cl2 = 1 g / 71 g/mol
Therefore, you would have approximately 0.032 mol of P and 0.014 mol of Cl2.
3. Next, determine the stoichiometry of the reaction. From the balanced equation, you can see that for every 2 moles of P, you need 3 moles of Cl2. This implies that the mole ratio of P to Cl2 is 2:3.
However, in this case, you have equal moles of P and Cl2, so whichever reactant runs out first will be the limiting reactant.
4. To determine the limiting reactant, compare the ratios of moles of each reactant to the stoichiometry. The limiting reactant is the one that produces fewer moles of the product (PCl3) in the given ratio.
- For P: 0.032 mol * (2/2) = 0.032 mol of PCl3 produced
- For Cl2: 0.014 mol * (2/3) = 0.0093 mol of PCl3 produced
Since Cl2 produces fewer moles of PCl3, it is the limiting reactant.
5. To find the excess reactant, subtract the moles of the limiting reactant from the moles of the other reactant.
Excess moles of P = moles of P - moles of PCl3 produced by P
= 0.032 mol - 0.032 mol
= 0 mol (none remaining)
Therefore, phosphorus is the excess reactant, and none of it will remain at the end of the reaction.
6. Finally, to determine the percentage of excess reactant remaining, divide the moles of the excess reactant by the initial moles of the excess reactant and multiply by 100%.
Percentage of P remaining = (moles of P remaining / initial moles of P) * 100%
= (0 mol / 0.032 mol) * 100%
= 0%
Therefore, the excess reactant is phosphorus (P), and none of it will remain at the end of the reaction.