Find # of g of each product thats produced from 100.0 g of Na2CO3 and 300. g of Fe3Br8 by reaction 4 Na2CO3 + Fe3Br8 8NaBr + 4CO2 + Fe3O4?

Molar mass of Na₂CO₃ = (22.99×2 + 12.01 + 16.00×3) g/mol = 105.99 g/mol
Molar mass of Fe₃Br₈ = (55.85×3 + 79.90×8) g/mol = 806.75 g/mol
Molar mass of NaBr = (22.99 + 79.90) g/mol = 102.89 g/mol
Molar mass of CO₂ = (12.01 + 16.0×2) g/mol = 44.01 g/mol
Molar mass of Fe₃O₄ = (55.85×3 + 16.00×4) g/mol = 231.55 g/mol

Initial number of moles of Na₂CO₃ = (100.0 g) / (105.99 g/mol) = 0.9435 mol
Initial number of moles of Fe₃Br₈ = (300.0 g) / (806.75 g/mol) = 0.3719 mol

4Na₂CO₃ + Fe₃Br₈ → 8NaBr + 4CO₂ + Fe₃O₄
Mole ratio Na₂CO₃ : Fe₃Br₈ = 4 : 1

When 0.9435 mol Na₂CO₃ completely reacts, Fe₃Br₈ needed = (0.9435 mol) × (1/4) = 0.2359 mol < 0.3719 mol
Fe₃Br₈ is in excess, and Na₂CO₃ is the limiting reactant/reagent.

According to the above equation, mole ratio Na₂CO₃ : NaBr : CO₂ : Fe₃O₄ = 4 : 8 : 4 : 1
Mass of NaBr produced = (0.9435 mol) × (8/4) × (102.89 g/mol) = 194.2 g
Mass of CO₂ produced = (0.9435 mol) × (4/4) × (44.01 g/mol) = 41.52 g
Mass of Fe₃O₄ produced = (0.9435 mol) × (1/4) × (231.55 g/mol) = 54.62 g