Sulfur trioxide, \(\mathrm{SO}_{3},\) is produced in enormous quantities each year for use in the synthesis of sulfuric acid. $$\mathrm{S}(s)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{SO}_{2}(g)$$ $$2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{SO}_{3}(g)$$ What volume of \(\mathrm{O}_{2}(g)\) at \(350 .^{\circ} \mathrm{C}\) and a pressure of 5.25 atm is needed to completely convert 5.00 \(\mathrm{g}\) sulfur to sulfur trioxide?

First, we need to convert the mass of sulfur (S) given in the problem (5.00 g) to moles. The molar mass of S is approximately 32.07 g/mol, so we have: Moles of S = \(\frac{5.00\, g}{32.07\, g\cdot mol^{-1}}\)

From the balanced chemical equation, we see that: - 1 mole of S reacts with 1 mole of O2 gas in the first reaction to produce SO2 (S + O2 → SO2). - In the second reaction (2 SO2 + O2 → 2 SO3), 2 moles of SO2 react with 1 mole of O2 gas to produce 2 moles of SO3. Because every mole of S reacts with a mole of O2 and we need one more mole of O2 for two moles of SO2, we will need 1.5 moles of O2 for every mole of S. Now we can find the moles of O2 needed: Moles of O2 = Moles of S × \(1.5 \, mol\, O_{2} \cdot mol^{-1}_{S}\)

Now that we have the moles of O2, we can use the ideal gas law to find the volume of O2 gas needed. The ideal gas law is expressed as: PV = nRT where P is the pressure, V is the volume, n is the number of moles, R is the gas constant, and T is the temperature. We are given the pressure (P = 5.25 atm) and temperature (T = 350°C = 623 K). We'll convert the given information to the proper units to use the Ideal Gas Law with the gas constant R = 0.0821 L·atm/mol·K. So, we can rearrange the ideal gas law equation to solve for V: V = \(\frac{nRT}{P}\) Now substitute the known values: V = \(\frac{(Moles\, of\, O_{2})(0.0821 L\cdot atm \cdot K^{-1} \cdot mol^{-1})(623\, K)}{5.25\, atm}\) Calculate the volume (V) with the obtained values, and you'll find the volume of O2 gas needed to completely convert 5.00 g of sulfur to sulfur trioxide.