To determine the mass of silver oxide produced from the reaction of oxygen gas with excess solid silver, we start by using the ideal gas law, which relates pressure, volume, and temperature to the number of moles of a gas. The formula for calculating moles is given by:

\( n = \frac{PV}{RT} \)

In this scenario, we are provided with the volume of oxygen gas (384 milliliters), the pressure (736 millimeters of mercury), and the temperature (25 degrees Celsius). First, we need to convert the pressure from millimeters of mercury to atmospheres, knowing that 1 atmosphere equals 760 millimeters of mercury:

\( P = \frac{736 \, \text{mmHg}}{760 \, \text{mmHg/atm}} = 0.9684 \, \text{atm} \)

Next, we convert the volume from milliliters to liters:

\( V = 384 \, \text{mL} = 0.384 \, \text{L} \)

We also need to convert the temperature to Kelvin by adding 273.15:

\( T = 25 \, \text{°C} + 273.15 = 298.15 \, \text{K} \)

Now, substituting these values into the ideal gas law equation, along with the universal gas constant \( R = 0.0821 \, \text{L} \cdot \text{atm} / (\text{mol} \cdot \text{K}) \), we can calculate the moles of oxygen gas:

\( n = \frac{(0.9684 \, \text{atm})(0.384 \, \text{L})}{(0.0821 \, \text{L} \cdot \text{atm} / (\text{mol} \cdot \text{K}))(298.15 \, \text{K})} \approx 0.015199 \, \text{mol} \)

With the moles of oxygen gas calculated, we can now perform a stoichiometric conversion to find the moles of silver oxide produced. The balanced chemical equation for the reaction is:

\( 2 \, \text{Ag} + \text{O}_2 \rightarrow 2 \, \text{Ag}_2\text{O} \)

This indicates that 1 mole of oxygen gas produces 2 moles of silver oxide. Therefore, we can set up the mole ratio:

\( \text{moles of } \text{Ag}_2\text{O} = 0.015199 \, \text{mol} \times \frac{2 \, \text{mol} \, \text{Ag}_2\text{O}}{1 \, \text{mol} \, \text{O}_2} = 0.030398 \, \text{mol} \, \text{Ag}_2\text{O} \)

Next, we convert moles of silver oxide to grams. The molar mass of silver oxide (\( \text{Ag}_2\text{O} \)) is calculated as follows:

\( \text{Molar mass of } \text{Ag}_2\text{O} = (2 \times 107.87 \, \text{g/mol}) + (1 \times 16.00 \, \text{g/mol}) = 230.74 \, \text{g/mol} \)

Now, we can find the mass of silver oxide produced:

\( \text{mass} = 0.030398 \, \text{mol} \times 230.74 \, \text{g/mol} \approx 7.0 \, \text{g} \)

In this calculation, we maintain two significant figures, as dictated by the initial temperature measurement. This process illustrates the application of stoichiometry in gas reactions, emphasizing the importance of understanding the relationships between moles, mass, and the ideal gas law.