We can use the ideal gas law, PV = nRT, to solve this problem.
First, we need to calculate the number of moles of SO2 that are produced:
PV = nRT
n = PV/RT
where P = 740 mmHg, V = 9.30 L, T = 125°C + 273.15 = 398.15 K, and R = 0.08206 L atm K^-1 mol^-1 is the ideal gas constant.
n = (740 mmHg) * (9.30 L) / (0.08206 L atm K^-1 mol^-1 * 398.15 K)
n = 0.356 mol
According to the balanced chemical equation for the combustion of sulfur to form sulfur dioxide:
S (s) + O2 (g) → SO2 (g)
one mole of sulfur reacts with one mole of oxygen to produce one mole of sulfur dioxide. Therefore, the number of moles of sulfur required is also 0.356 mol.
To calculate the mass of sulfur that must react, we need to use the molar mass of sulfur:
M(S) = 32.06 g/mol
mass of sulfur = number of moles of sulfur * molar mass of sulfur
mass of sulfur = 0.356 mol * 32.06 g/mol
mass of sulfur = 11.43 g
Therefore, 11.43 g of sulfur must react to produce 9.30 L of sulfur dioxide at 740 mmHg and 125°C.