12.4 x 10^40 molecules of butane combust w 12.4L of O2 at stp find g of H2O produced?

2C₄H₁₀(g) + 13O₂(g) → 8CO₂(g) + 10H₂O(l)

Initial no. of moles of C₄H₁₀ = (12.4 × 10⁴⁰) / (6.022 × 10²³ /mol) = 2.06 × 10¹⁷ mol
Initial no. of moles of O₂ = (12.4 L) / (22.4 L/mol) = 0.5536 mol
Obviously, O₂ is the limiting reactant (completely reacts).

Refer to the equation. 13 moles of O₂ reacts to give 10 moles of H₂O.
No. of moles of H₂O produced = (0.5536 mol) × (10/13) = 0.4258 mol

Molar mass of H₂O = 1.01×2 + 16.00 = 18.02 g/mol
Mass of H₂O produced = (18.02 g/mol) × (0.4258 mol) = 7.67 g