what is the percentage yield of a reaction in which 200 gram PCl3 react with excess water to form 128 of HCl according to the following reaction. PCl3 + 3H2O it gives H3PO3 + 3HCl.
First, we need to determine the balanced chemical equation:
PCl3 + 3H2O → H3PO3 + 3HCl
Next, we'll find the molar mass of each compound:
PCl3: P = 30.97 g/mol, Cl = 35.45 g/mol
Molar mass of PCl3 = 30.97 + 3 * 35.45 = 137.33 g/mol
HCl: H = 1.01 g/mol, Cl = 35.45 g/mol
Molar mass of HCl = 1.01 + 35.45 = 36.46 g/mol
Now we'll calculate the moles of PCl3 and HCl:
moles of PCl3 = mass / molar mass = 200 g / 137.33 g/mol = 1.456 moles
According to the balanced chemical equation, 1 mole of PCl3 produces 3 moles of HCl:
moles of HCl produced = 1.456 moles PCl3 * 3 = 4.368 moles
Now we'll convert moles of HCl back to grams:
mass of HCl produced = 4.368 moles * 36.46 g/mol = 159.20 g
We're given that the actual mass of HCl formed is 128 g, so we can calculate the percentage yield:
percentage yield = (actual mass / theoretical mass) * 100%
percentage yield = (128 g / 159.20 g) * 100% = 80.4%
The percentage yield of the reaction is 80.4%.