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? What volume of water (d= 1.00 g/cm3) should be?

To calculate the amount of phosphorous triiodide (PI3) that should be weighed out and the volume of water required, we need to follow these steps:

Step 1: Determine the molar mass of phosphorous triiodide (PI3)
The molar mass of phosphorous (P) is 30.97 g/mol, and the molar mass of iodine (I) is 126.90 g/mol. Since phosphorous triiodide (PI3) consists of one phosphorous atom (P) and three iodine atoms (I), we can calculate the molar mass as follows:

Molar mass of PI3 = (1 * molar mass of P) + (3 * molar mass of I)
Molar mass of PI3 = (1 * 30.97 g/mol) + (3 * 126.90 g/mol)
Molar mass of PI3 = 411.67 g/mol

Step 2: Calculate the mass of phosphorous acid required
Since we need to obtain a volume of 0.250 L of phosphorous acid with a density of 1.651 g/mL, we can use the following equation:

Mass = Volume * Density
Mass = 0.250 L * 1.651 g/mL
Mass = 0.413 g

Step 3: Adjust for the yield
The given procedure mentions a yield of 75.0%, which means only 75.0% of the desired amount of phosphorous acid will be obtained. Therefore, we need to adjust the mass calculated in step 2 as follows:

Adjusted mass = Mass / Yield
Adjusted mass = 0.413 g / 0.75
Adjusted mass = 0.551 g

Step 4: Determine the mass of phosphorous triiodide required
Since the reaction between phosphorous triiodide (PI3) and water (H2O) has a 45.0% excess of water, we need to account for this excess by considering the molar ratio between PI3 and H2O. From the balanced equation, we know that the ratio of PI3 to H2O is 1:1:

Mass of PI3 required = Adjusted mass / (1 - % excess)
Mass of PI3 required = 0.551 g / (1 - 0.45)
Mass of PI3 required = 1.002 g

Therefore, the student should weigh out approximately 1.002 g of phosphorous triiodide.

Step 5: Determine the volume of water required
The volume of water required can be calculated using the mass of water required and the density of water. Given that the density of water is 1.00 g/cm3 and the mass of water needed:

Volume = Mass / Density
Volume = 1.002 g / 1.00 g/cm3
Volume = 1.002 cm3

Since the density of water and the volume units are the same, the volume of water required is approximately 1.002 cm3.

Therefore, the student should weigh out approximately 1.002 g of phosphorous triiodide and use approximately 1.002 cm3 of water for the reaction.