6.3 liters of sulfur vapor, S8(g), at 655°C and 8.87 atm is burned in excess pure oxygen gas to give sulfur dioxide gas measured at the ...?

Method 1:

Consider the S₈ vapor reacted:
Pressure, P = 8.87 atm
Volume, V = 6.3 L
Temperature, T = (273 + 655) K = 928 K
Gas constant, R = 0.08206 L atm / (mol K)

Gas law: PV = nRT
Then, n = PV/(RT)

Moles of S₈ reacted, n = 8.87 × 6.3 / (0.08206 × 928) mol = 0.7338 mol
Moles of SO₂ obtained = 0.7338 × 8 = 5.8704 mol

Molar mass of SO₂ = (32.1 + 16.0×2) g/mol = 64.1 g/mol
Mass of SO₂ obtained = (5.8704 mol) × (64.1 g/mol) = 376.3 g

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Method 2:

At constant temperature and pressure, volume ratio of gases is equal to their mole ratio.

Balanced equation for the reaction:
S₈ + 8O₂ → 8SO₂
Volume ratio S₈ : SO₂ = 1 : 8
Volume of SO₂ obtained = (6.3 L) × 8 = 50.4 L

Consider the SO₂ gas obtained:
Pressure, P = 8.87 atm
Volume, V = 50.4 L
Temperature, T = (273 + 655) K = 928 K
Molar mass, M = (32.1 + 16.0×2) g/mol = 64.1 g/mol
Gas constant, R = 0.08206 L atm / (mol K)

PV = nRT and n = m/M
Then, PV = (m/M)RT
m = PVM/(RT)

Mass of SO₂ produced, m
= 8.87 × 50.4 × 64.1 / (0.08206 × 928) g
= 376.3 g

1 S8 + 8 O2 --> 8 SO2
6.3L S8 * (8L SO2 / 1L S8) = 50.4L SO2
PV = nRT --> PV = (mass / mw) RT --> mass = mw * PV/(RT)

you get to finish up.