You are working in my lab. I ask you to make 0.250 L of H3PO4 (density is 1.651 g/cm³). I give you solid phosphorus triiodide with water. The other product will be hydrogen iodide gas. I tell you to use 45.0 % excess water and that the reaction will yield 75.0 % H3PO4. How much phosphorus triiodide should be weighed out?

Note: Phosphorus triiodide (PI3) actually reacts with water (H2O) to form phosphorous acid, H3PO3 whose density is 1.651 g/cm3 or 1.651 g/mL, and not H3PO4. This could be a typo in the question. The balanced equation is: PI3(s) + 3H2O(l) -----> H3PO3(aq) + 3HI(g) Since water (H2O) is used in excess, the limiting reactant is Phosphorus triiodide (PI3).Given the volume of H3PO3 that need to be prepared, V = 0.250 L * 1000 mL1 L= 250 mL density of H3PO3, d = 1.651 g/mL Conversion factor: 1 mL = 1.651 g Mass of H3PO3 that need to be prepared, m = V*d= 250 mL * 1.651 g1 mL= 412.75 g So the actual yield = 412.75 g Given the percent yield of H3PO3 = 75.0 % The formula for percent yield is: Percent Yield = Actual yieldTheoretical yield*100⇒75.0 = 412.75 g Theoretical yield*100⇒Theoretical yield = 412.75 g 75.0*100⇒Theoretical yield = 550.3 g Theoretical yield of H3PO3 = 550.3 g Molar mass of H3PO3 = 82.0 g/mol Theoretical moles of H3PO3 required = 550.3 g * 1 mol82.0 g= 6.711382 mol In the balanced equation, the mole ratio between H3PO3 and PI3(s) is 1/1 Hence theoretical moles of H3PO3 = theoretical moles of PI3(s) = 6.711382 mol Molar mass of PI3 = 411.69 g/mol Mass of Phosphorus triiodide (PI3) required = 6.711382 mol * 411.69 g1 mol= 2763 g (Answer)