Answer:
mass = 172.78 grams Na₂CO₃(s) formed
Step-by-step explanation:
6Na°(s) + Fe₂(CO₃)₃ => 3Na₂CO₃(s) + 2Fe°(s)
moles Na°(s) = 15g/23g/mol = 3.26 mole Na°(s)
From stoichiometry of reaction equation, 3.26 mole Na°(s) => 3/6(3.26) mole Na₂CO₃(s) = 1.63 mole Na₂CO₃(s) x 106 g/mole = 172.78 grams Na₂CO₃(s)