Given the equation: H2 + I2 -> 2HI?

Molar mass of H₂ = 1.0 × 2 g/mol = 2.0 g/mol
Molar mass of I₂ = 127 × 2 g/mol = 254 g/mol

H₂(g) + I₂(g) → 2HI(g)
Mole ratio H₂ : I₂ = 1 : 1

Initial number of moles of H₂ = (62.0 g) / (2.0 g/mol) = 31.0 mol
Initial number of moles of I₂ = (62.0 g) / (254 g/mol) = 0.244 mol < 31.0 mol
Hence, I₂ completely reacts.

According to the above equation, mole ratio I₂ : HI = 1 : 2
Number of moles of I₂ reacted = 0.244 mol
Number of moles of HI produced = (0.244 mol) × 2 = 0.488 mol

For the HI produced :
Pressure, P = 752/760 atm
Number of moles, n = 0.488 mol
Gas constant, R = 0.08206 L atm / (mol K)
Temperature, T = (273.2 + 14.0) K = 287.2 K
Volume, V = ? L

Gas law: PV = nRT
Hence, V = nRT/P

Volume of HI produced, V = 0.488 × 0.08206 × 287.2 / (752/760) L = 11.6 L