Answer to Question #337540 in General Chemistry for Marianne Grace

Question

Thermite reaction is necessary to produce molten iron. this reaction involves Fe3O4 and Al and is described by:

8 Al + 3 Fe3O4 --> 9 Fe + 4 Al2O3

If 198.5 g Al and 365.5 g Fe3O4 are used, what mass of the excess reagent remains unused at the end of the reaction?

Answer 1

8Al + 3Fe₃O₄ --> 9Fe + 4Al₂O₃

n(Al) = m(Al)/8Mr(Al) = 198.5 g / 8*26.98 g/mol = 0.92 moles

n(Fe₃O₄) = m(Fe₃O₄)/3Mr(Fe₃O₄) = 365.5 g / 3*231.53 g/mol = 0.53 moles

Al is the excess

by equation:

m(Al) = (m(Fe₃O₄) * 8Mr(Al)) / 3Mr(Fe₃O₄)

m(Al) = (365.5 g * 8*26.98 g/mol) / 3*231.53 g/mol = 113.58 g

Therefore the mass of the excess reagent that remains 

m(Al) = 198.5 g - 113.58 g = 84.92 g