A student prepares phosphorous acid by reacting solid phosphorous triodide with water

PI3+H2O-->H3PO3+HI

The student needs to obtain 0.250L of phosphorous acid. (d=1.651g/mL). The procedure calls for a 45.0% excess of water and a yield of 75.0%. How much phosphorous triiodide should be weighed out?

My brain is fried from math today can someone help with the steps to solve this???

Balance the equation.

PI3 + 3H2O ==> H3PO3 + 3HI

I believe the problem assumes the H3PO3 prepared in this manner will be 100% purity; therefore, how many moles is in the product?
1.651 g/mL x 250 mL = 412.75 grams.
412.75/molar mass H3PO3 = 5.034 moles H3PO3.
The amount of PI3 must also be 5.034 moles (since 1 mole PI3 produces 1 mole H3PO3). Convert that to grams PI3. Since it is only 75% efficient, divide the final # grams by 0.75.

To solve this problem, we can break it down into several steps:

Step 1: Determine the desired amount of phosphorous acid.
The student needs to obtain 0.250 L of phosphorous acid.

Step 2: Calculate the mass of phosphorous acid.
Since we know the density of phosphorous acid is 1.651 g/mL, we can calculate the mass of phosphorous acid using the formula:
Mass = Volume x Density
Mass = 0.250 L x 1.651 g/mL = 0.41275 g

Step 3: Determine the amount of water needed.
Since the student needs to use a 45.0% excess of water, we need to calculate the amount of water required for the reaction.
Excess water = 45.0% of the amount of water needed for the reaction = 0.45 x 0.250 L = 0.1125 L

Step 4: Calculate the amount of phosphorous acid produced.
Since the equation is balanced as follows:
1 mol PI3 + 3 mol H2O → 1 mol H3PO3 + 3 mol HI
From the balanced equation, we can see that 1 mole of PI3 produces 1 mole of H3PO3.
However, the reaction only has a yield of 75.0%, so the amount of phosphorous acid actually produced is:
Amount of H3PO3 produced = 75.0% of the desired amount of phosphorous acid = 0.75 x 0.41275 g = 0.30956 g

Step 5: Calculate the amount of PI3 needed.
Since 1 mole of PI3 produces 1 mole of H3PO3, and the molar mass of PI3 is 411.7 g/mol, we can calculate the amount of PI3 needed using the formula:
Amount of PI3 = Amount of H3PO3 produced / Molar mass of PI3
Amount of PI3 = 0.30956 g / 411.7 g/mol = 0.000750 mol

Step 6: Convert moles to grams.
To calculate the mass of PI3 needed, use the formula:
Mass of PI3 = Amount of PI3 x Molar mass of PI3
Mass of PI3 = 0.000750 mol x 411.7 g/mol = 0.309525 g

Therefore, the student should weigh out approximately 0.309525 grams of phosphorous triiodide.

To solve this problem, we need to follow a step-by-step approach.

Step 1: Determine the volume of phosphorous acid needed.
The student needs 0.250 L of phosphorous acid.

Step 2: Calculate the mass of phosphorous acid using its density.
Given that the density of phosphorous acid is 1.651 g/mL, we can use the formula:
mass = volume × density
mass = 0.250 L × 1.651 g/mL

Step 3: Calculate the mass of phosphorus triiodide (PI3) required.
The balanced equation provides the molar ratio between PI3 and H3PO3. Comparing the coefficients, we find that one mole of PI3 is required to produce one mole of H3PO3. Thus, the molar ratio is 1:1.

To calculate the mass of PI3 needed, we need to convert the desired mass of H3PO3 (from Step 2) to moles, and then convert moles of H3PO3 to moles of PI3 by applying the molar ratio.

Step 3.1: Convert the mass of H3PO3 to moles.
Using the molar mass of H3PO3, which is 97.99 g/mol (according to the periodic table), we can convert the mass of H3PO3 to moles:
moles of H3PO3 = mass of H3PO3 / molar mass of H3PO3

Step 3.2: Calculate the moles of PI3 required using the molar ratio.
Since the molar ratio is 1:1 between PI3 and H3PO3, the moles of PI3 required will be the same as moles of H3PO3

Step 3.3: Convert moles of PI3 to mass.
To convert moles of PI3 to mass, we will use the molar mass of PI3, which is 411.69 g/mol (according to the periodic table):
mass of PI3 = moles of PI3 × molar mass of PI3

Step 4: Adjust for a 75.0% yield.
The yield mentioned in the problem is 75.0%. To account for the yield, we need to divide the calculated mass by 0.75:
adjusted mass of PI3 = mass of PI3 / 0.75

Step 5: Calculate the mass of PI3 required, including a 45.0% excess of water.
For each mole of PI3, we need 1 mole of H2O. Thus, to calculate the mass of PI3 required, we need to consider the molar mass of H2O and the percentage excess of water:
mass of PI3 required = adjusted mass of PI3 + (0.45 × molar mass of H2O)

Now, you can plug in the values and solve for the mass of PI3 required.