How many grams of PI3 could actually be produced from 250 g of I2 and excess phosphorous if the reaction gives a 98.5% yield?
P4 + 6I2 --> 4PI3
Well, I could give you a serious answer, but I think I'll clown around a bit. So, in this fun little reaction, 1 mole of P4 reacts with 6 moles of I2 to produce 4 moles of PI3.
Now, knowing that the molar mass of I2 is approximately 253.8 g/mol and the molar mass of PI3 is roughly 411.8 g/mol, we can calculate how much PI3 could be produced.
First, let's find out how many moles of I2 we have. By dividing 250 g by the molar mass of I2, we get approximately 0.985 moles. Since there is an excess of phosphorus, we only need to consider the limiting reagent, which is I2 in this case.
Now, multiplying the moles of I2 by the mole ratio of PI3 to I2 (4:6), we find that we could produce approximately 0.6567 moles of PI3.
Finally, we can calculate the mass of PI3 by multiplying the moles of PI3 by its molar mass. Doing the math, we get approximately 269.7 grams of PI3 could be produced.
But remember, this calculation assumes a 98.5% yield, so it's like telling a joke with a punchline that might not hit the mark.
To determine the number of grams of PI3 that could be produced from 250 g of I2, we first need to calculate the molar mass of I2 and PI3.
The molar mass of I2 (iodine) is 2 × 126.90447 g/mol = 253.80894 g/mol.
The molar mass of PI3 (phosphorus triiodide) is (1 × 30.973762 g/mol) + (3 × 126.90447 g/mol) = 411.68892 g/mol.
Now we can set up the stoichiometry of the reaction:
1 mol P4 + 6 mol I2 -> 4 mol PI3
From the balanced equation, we can see that for every 6 moles of I2, we can produce 4 moles of PI3.
Next, we will convert the mass of I2 to moles by dividing the mass by the molar mass:
moles of I2 = 250 g / 253.80894 g/mol = 0.9858 mol
Since the reaction has a 98.5% yield, we can multiply the number of moles of I2 by the yield to obtain the actual moles of PI3 produced:
moles of PI3 = 0.9858 mol × 0.985 = 0.9702 mol
Finally, we can calculate the mass of PI3 by multiplying the moles by the molar mass:
mass of PI3 = 0.9702 mol × 411.68892 g/mol = 399.1 g
Therefore, approximately 399.1 grams of PI3 could be produced from 250 grams of I2 with a 98.5% yield.
To determine the number of grams of PI3 that can be produced, we need to use stoichiometry and the concept of limiting reagents.
Step 1: Write and balance the equation for the reaction:
P4 + 6I2 -> 4PI3
Step 2: Convert the given mass of I2 to moles:
molar mass of I2 = 127.9 g/mol
moles of I2 = mass of I2 / molar mass of I2
moles of I2 = 250 g / 127.9 g/mol
Step 3: Use the stoichiometric ratio to find the moles of PI3 produced. From the balanced equation, we see that 1 mole of P4 produces 4 moles of PI3:
moles of PI3 = moles of I2 / 6 * (4/1)
Step 4: Calculate the mass of PI3 produced:
molar mass of PI3 = 137.9 g/mol
mass of PI3 = moles of PI3 * molar mass of PI3
Step 5: Apply the yield percentage to calculate the actual amount of PI3 obtained:
actual mass of PI3 = mass of PI3 * yield percentage
Plug in the known values to calculate the answer.
Note: When calculating the answer, it is important to use the correct number of significant figures based on the given values and the precision required by the question.
How many mols I2 do you have? mols = g/molar mass
4 moles I2 produce 6 mols PI3.
How many grams is that? g = moles x molar mass That give you the theoretical yield. Convert to actual yield by theoretical yield x 0.985 = actual yield.
By the way, you misspelled phosphorus.