Final answer:
To determine the volume of water produced when 250 L of hydrogen gas are used to reduce copper (II) oxide at STP, we use stoichiometry and the ideal gas law.
The volume of water produced is 259 L.
Step-by-step explanation:
To determine the volume of water produced when 250 L of hydrogen gas are used to reduce copper (II) oxide at STP, we need to use stoichiometry and the ideal gas law. First, we calculate the number of moles of hydrogen using the ideal gas law: PV = nRT. At STP, the pressure is 1 atm and the temperature is 273 K, so we have:
n = PV / RT = (1 atm x 250 L) / (0.0821 L·atm/mol·K x 273 K) = 11.47 moles of H2
Now, we use the balanced chemical equation to determine the stoichiometric ratio between hydrogen and water: 2 moles of H2 produces 2 moles of H2O. Therefore, 11.47 moles of H2 will produce 11.47 moles of H2O. Finally, we convert moles of water to volume using the ideal gas law again: V = nRT / P = (11.47 moles x 0.0821 L·atm/mol·K x 273 K) / 1 atm = 259 L of water.