Answer to Question #99832 in General Chemistry for Ky

Question

what is the theoretical yield of iron oxide when 55.65g of iron (ii) nitrate reacts with 12.75g of aluminum oxide?

Answer 1

Reaction: 3Fe(NO₃)₂ + Al₂O₃ = 3FeO + 2Al(NO₃)₃

Amount of substance: n(Fe(NO₃)₂) = m/M = 55,65/180 = 0,309 mol

n(Al₂O₃) = 12,75/102 = 0,125 mol

So, the Al₂O₃ is in excess. Therefore, n(Fe(NO₃)₃)=n(FeO)

So, the yeild of FeO: m = n•M = 0,309•72 = 22,25 g.