Final answer:
Approximately 1.52 liters of hydrogen gas is produced when 1.52 g of magnesium reacts with hydrochloric acid at 23°C and 1.00 atm.
Step-by-step explanation:
The student wants to know the volume of hydrogen gas produced when 1.52 g of solid magnesium reacts with hydrochloric acid (HCl) at a specified temperature and pressure.
Solution
First, calculate the number of moles of magnesium:
Molar mass of Mg = 24.305 g/mol
Moles of Mg = mass (g) / molar mass (g/mol) = 1.52 g / 24.305 g/mol ≈ 0.0625 mol
Using the stoichiometry of the reaction, Mg(s) + 2 HCl(aq) → MgCl₂(aq) + H₂(g), each mole of Mg produces one mole of H₂. Therefore, 0.0625 mol of Mg will produce 0.0625 mol of H₂.
Using the ideal gas law, PV = nRT, and converting temperature to Kelvin (T=23°C+273=296K), we can solve for V (volume of H₂ at 1 atm and 296 K).
R (ideal gas constant) = 0.0821 L·atm/mol·K
V = nRT/P
V = (0.0625 mol) * (0.0821 L·atm/mol·K) * (296 K) / (1.00 atm)
V ≈ 1.52 L
Therefore, the volume of hydrogen gas produced is approximately 1.52 liters.