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???

Sure, I can help you with that. To solve this problem, we need to follow a few steps:

Step 1: Convert the given volume of phosphorous acid to its mass:
Given volume = 0.250 L
Density of phosphorous acid = 1.651 g/mL
Mass = Volume x Density
Mass = 0.250 L x 1.651 g/mL = 0.41275 g

Step 2: Calculate the mass of phosphorous triiodide needed (assuming a 75.0% yield):
Mass of phosphorous triiodide = Mass of phosphorous acid / Yield
Mass of phosphorous triiodide = 0.41275 g / 0.75 = 0.55033 g

Step 3: Determine the molar mass of phosphorous triiodide (PI3):
Phosphorous (P) atomic mass = 31 g/mol
Iodine (I) atomic mass = 127 g/mol
Molar mass of PI3 = (P atomic mass x 1) + (I atomic mass x 3)
Molar mass of PI3 = (31 g/mol x 1) + (127 g/mol x 3) = 412 g/mol

Step 4: Calculate the number of moles of phosphorous triiodide:
Number of moles = Mass / Molar mass
Number of moles = 0.55033 g / 412 g/mol = 0.0013354 mol

Step 5: Determine the amount of water needed as a 45.0% excess:
Water excess = 45.0% of the number of moles of phosphorous triiodide
Water excess = 0.45 x 0.0013354 mol = 0.0006009 mol

Step 6: Determine the number of moles of water needed:
Number of moles of water = Number of moles of phosphorous triiodide + Water excess
Number of moles of water = 0.0013354 mol + 0.0006009 mol = 0.0019363 mol

Step 7: Convert the number of moles of water to its mass:
Mass of water = Number of moles x Molar mass of water
Mass of water = 0.0019363 mol x 18 g/mol (molar mass of water) = 0.0348534 g

Step 8: Calculate the mass of phosphorous triiodide required, considering the excess water:
Mass of phosphorous triiodide = Mass of phosphorous triiodide needed - Mass of water
Mass of phosphorous triiodide = 0.55033 g - 0.0348534 g = 0.51548 g

Therefore, the student should weigh out approximately 0.51548 g of phosphorous triiodide.