how much acetylene will be produced when 500g of calcium carbide reacts with 1 gallon of water?

Molar mass of CaC₂ (calcium carbide) = (40.1 + 12.0×2) g/mol = 64.1 g/mol
Molar mass of H₂O (water) = (1.0×2 + 16.0) g/mol = 18.0 g/mol
Molar mass of C₂H₂ (acetylene) = (12.0×2 + 1.0×2) g/mol = 26.0 g/mol

Volume of water = 1 galloon = 3.785 liter = 3785 mL
Density of water = 1.00 g/mL
Mass of water = (3785 mL) × (1 g/mL) = 3785 g

CaC₂ + 2H₂O → Ca(OH)₂ + C₂H₂

Initial number of moles of CaC₂ = (500 g) / (64.1 g/mol) = 7.80 mol
Initial number of moles of H₂O = (3785 g) / (18.0 g/mol) = 210 mol ≫ 7.8 mol
Obviously, H₂O is in largely excess, and CaC₂ completely reacts.

According to the above equation, mole ratio CaC₂ : C₂H₂ = 1 : 1
Number of moles of CaC₂ reacted = 7.80 mol
Number of moles of C₂H₂ produced = 7.80 mol
Mass of C₂H₂ produced = (7.80 mol) × (26.0 g/mol) = 203 g (3 sig. fig.)