Calculate the volume of H₂(g) at 273 K and 2.00 atm that will be formed when 275 mL of 0.725 M HCl solution reacts with 50.0 g Zn(s) to give hydrogen gas and aqueous zinc chloride. (R = 0.08206 L • atm/K • mol) A) 0.56 L B) 1.12 L C) 2.23 L D) 4.47 L E) 3.54 L

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First, calculate moles of HCl: 0.275 L * 0.725 mol/L = 0.199 mol HCl. The reaction is Zn + 2HCl → ZnCl2 + H2, so 0.199 mol HCl produces 0.0995 mol H2. Calculate moles of Zn: 50.0 g / 65.38 g/mol = 0.765 mol Zn, which is excess. Use PV = nRT to find volume: V = (0.0995 mol * 0.08206 L atm/mol K * 273 K) / 2.00 atm = 1.12 L.

Calculate the Volume of H2(g) at 273 K and 2

Calculate the Volume of H₂(g) at 273 K and 2