Answer to Question #291386 in Chemistry for LINDA

Question

IF YOU START WITH 50N G OF IRON III OXIDE,AND 25 G AL, WHAT IS THEMAXIMUM MASS OF ALUMINUM OXIDE THAT COULD BE PRODUCED? HOW MUCH AL OXIDE WOULDNBE PRODUCED IF HE YIELD IS 93%

Answer 1

Fe₂O₃ + 2Al = Al₂O₃ + 2Fe

According to the equation, n (Al₂O₃) = 2 n (Al) = n (Fe₂O₃)

M (Fe₂O₃) = 159.7 g/mol

M (Al) = 26.98 g/mol

M (Al₂O₃) = 101.96 g/mol

n (Fe₂O₃) = m / M = 50 / 159.7 = 0.31 mol

n (Al) = 25 / 26.98 = 0.93 mol

Fe₂O₃ is the limiting reactant.

n (Al₂O₃) = n (Fe₂O₃) = 0.31 mol

m (Al₂O₃) = n x M = 0.31 x 101.96 = 31.61 g

m (Al₂O₃)_93% = 31.61 x 0.93 = 29.4 g