If 5.00 g of PCl5 is reacted with excess water, how many grams of H3PO4 and HCl would be produced?
If the titre value of the experiment were 20.50, 22.45, 22.6, typical experiment in a laboratory, 4.50g of sulphuric acid in 250cm^3 of water was titrated with an impure solution of 5.24g of sodium hydroxide contained in 1dm^3.
To determine the grams of H3PO4 and HCl produced, we need to understand the balanced chemical equation for the reaction between PCl5 and water. The balanced equation is as follows:
PCl5 + 4H2O → H3PO4 + 5HCl
The stoichiometry of the equation tells us that 1 mole of PCl5 reacts with 4 moles of water to produce 1 mole of H3PO4 and 5 moles of HCl.
To solve the problem, we will use the given amount of PCl5 (5.00 g) to find the number of moles of PCl5. Then, using the stoichiometric ratios, we can determine the number of moles of H3PO4 and HCl produced. Finally, we will convert the moles back into grams.
1. Calculate the moles of PCl5:
Molar mass of PCl5 = 31.0 g/mol (atomic mass of P = 31.0 g/mol and atomic mass of Cl = 35.5 g/mol)
Moles of PCl5 = Mass (g) / Molar mass (g/mol)
Moles of PCl5 = 5.00 g / 208.5 g/mol ≈ 0.024 moles
2. Use the stoichiometric ratio:
From the balanced equation: 1 mole of PCl5 produces 1 mole of H3PO4.
Moles of H3PO4 = 0.024 moles
From the balanced equation: 1 mole of PCl5 produces 5 moles of HCl.
Moles of HCl = 5 * 0.024 moles = 0.120 moles
3. Convert moles back to grams:
Molar mass of H3PO4 = 98.0 g/mol
Molar mass of HCl = 36.5 g/mol
Mass of H3PO4 = Moles of H3PO4 * Molar mass of H3PO4
Mass of H3PO4 = 0.024 moles * 98.0 g/mol ≈ 2.35 g
Mass of HCl = Moles of HCl * Molar mass of HCl
Mass of HCl = 0.120 moles * 36.5 g/mol ≈ 4.38 g
Therefore, approximately 2.35 grams of H3PO4 and 4.38 grams of HCl would be produced when 5.00 grams of PCl5 is reacted with excess water.