Final answer:
The mass of magnesium required to produce 225 ml of hydrogen gas at STP is 0.244 g, which does not match any of the provided answer choices. None of the Option is correct.
Step-by-step explanation:
To determine the mass of magnesium metal that reacts with sulfuric acid to produce 225 ml of hydrogen gas at STP (standard temperature and pressure), we must use stoichiometry and the ideal gas law. At STP, 1 mole of gas occupies 22.4 liters. Hence, we can calculate the number of moles of hydrogen gas:
Convert the volume of H₂ from milliliters to liters: 225 ml = 0.225 L.
Calculate the number of moles of H₂ produced using the molar volume at STP: Moles H₂ = 0.225 L / 22.4 L/mol = 0.01004 mol.
The balanced equation for the reaction of magnesium with sulfuric acid is: Mg(s) + H₂SO₄(aq) → MgSO₄(aq) + H₂(g).
From the balanced equation, 1 mole of Mg produces 1 mole of H₂. Therefore, the number of moles of Mg needed is also 0.01004 mol.
We then calculate the mass of Mg: Mass = moles × molar mass = 0.01004 mol × 24.305 g/mol = 0.244 g.
However, none of the answers provided (A) 0.63 g, (B) 1.26 g, (C) 1.89 g, (D) 2.52 g match the calculated mass. It's possible there could be an error in the question or answer choices provided. Based on the calculation, the mass should be approximately 0.244 g, which is not an available answer choice.