What volume of H2 will be released following the reaction between 40 g Zn (65.41 g mol-1) and excess HCl if the gas is collected at 400 K at a pressure of 2 atmospheres. R = 0.08205 L atm mol-1 K-1

To find the volume of hydrogen gas (H2) released when 40 grams of zinc (Zn) react with excess hydrochloric acid (HCl) at 400 K and 2 atmospheres of pressure, you can use the ideal gas law: PV = nRT Where: P = pressure (in atm) V = volume (in liters) n = moles of gas R = ideal gas constant (0.08205 L·atm/(mol·K)) T = temperature (in Kelvin) First, calculate the moles of zinc (Zn) used in the reaction: 1. Calculate the molar mass of zinc (Zn): Molar mass of Zn = 65.41 g/mol 2. Calculate the moles of Zn used: Moles of Zn = Mass / Molar mass Moles of Zn = 40 g / 65.41 g/mol ≈ 0.611 mol The balanced chemical equation for the reaction between zinc (Zn) and hydrochloric acid (HCl) is: Zn + 2HCl -> ZnCl2 + H2 From the balanced equation, you can see that 1 mole of Zn produces 1 mole of H2 gas. So, 0.611 moles of Zn will produce 0.611 moles of H2 gas. Now, you can use the ideal gas law to calculate the volume of H2 gas at the given conditions: P = 2 atm T = 400 K PV = nRT V = (nRT) / P V = (0.611 mol * 0.08205 L·atm/(mol·K) * 400 K) / 2 atm V ≈ 6.16 liters Therefore, approximately 6.16 liters of hydrogen gas (H2) will be released when 40 grams of zinc (Zn) react with excess hydrochloric acid (HCl) at 400 K and 2 atmospheres of pressure.