Final answer:
To calculate the grams of Ag₂O needed to produce 3.40 L of oxygen gas at given conditions, the ideal gas law and stoichiometry are used to first find the moles of oxygen, then the moles of Ag₂O, which are finally converted to grams.
Step-by-step explanation:
The subject in question is Chemistry, specifically pertaining to the decomposition reactions and calculations using the ideal gas law. To determine how many grams of Ag₂O are required to produce 3.40 L of oxygen gas at a pressure of 102.3 kPa and a temperature of 175°C, we must first use the ideal gas equation PV = nRT to find the moles of oxygen gas.
Let's utilize the known values: P = 102.3 kPa which converts to 1.01 atm (using 101.3 kPa/atm), V = 3.40 L, R = 0.0821 L·atm/K·mol, and T = 175°C which is 448.15 K (using T(K) = T(°C) + 273.15). Solving for n (moles of O₂) gives us n = PV/RT.
After calculating the moles of O₂, the balanced equation 2Ag₂O → 4Ag + O₂ tells us that 2 moles of Ag₂O produce 1 mole of O₂. With this stoichiometric ratio, we can determine the required moles of Ag₂O to produce the calculated moles of O₂. Finally, with the molar mass of Ag₂O, we can convert moles to grams.