what mass of water is needed to react with 10.0 grams of PI3?
To determine the mass of water needed to react with 10.0 grams of PI3, we first need to understand the balanced chemical equation for the reaction between PI3 (phosphorus triiodide) and water.
The balanced chemical equation for the reaction is as follows:
PI3 + 3H2O -> H3PO3 + 3HI
From the balanced equation, we can see that one mole of PI3 reacts with three moles of water to produce one mole of H3PO3 (phosphorous acid) and three moles of HI (hydroiodic acid).
Now, in order to find the mass of water needed, we need to convert the given mass of PI3 into moles. To do this, we need to know the molar mass of PI3, which can be calculated by summing the atomic masses of its constituent elements: phosphorus (P) and three iodine atoms (I).
The atomic mass of phosphorus (P) is approximately 31.0 g/mol, and the atomic mass of iodine (I) is approximately 126.9 g/mol. Adding these values gives us a molar mass of PI3 of approximately 31.0 g/mol + (3 x 126.9 g/mol) = 411.7 g/mol.
Next, we can calculate the number of moles in 10.0 grams of PI3 by dividing the given mass by the molar mass:
10.0 g PI3 / 411.7 g/mol = 0.0243 mol PI3
Since the stoichiometric ratio between PI3 and water is 1:3, we know that 0.0243 moles of PI3 will react with 3 x 0.0243 moles of water.
Therefore, the mole ratio tells us that 0.0243 moles of PI3 will react with 3 x 0.0243 moles of water, which is equal to 0.0729 moles of water.
Finally, to find the mass of water needed, we can multiply the number of moles of water by its molar mass. The molar mass of water (H2O) is approximately 18.0 g/mol:
0.0729 mol H2O x 18.0 g/mol = 1.312 g
Therefore, approximately 1.312 grams of water is needed to react with 10.0 grams of PI3.