Question

1. Given: 2Al(s) + Fe2O3(s) à Al2O3(s) + 2Fe(l) Step 1 is done!
How many grams of Fe2O3 are needed to produce 6.7 g of Fe?
Step 2: Mass to moles
Step 3: Mole to Mole (Ratio)
Step 4: Mole to Mass

Answer 1

2Al_(s) + Fe₂O_3(s) = Al₂O_3(s) + 2Fe(l)

n=m/M

M (Fe) = 55.9 g/mol

M (Fe₂O₃) = 159.7 g/mol

n (Fe)=6.7/55.9 = 0.12 mol

n (Fe₂O₃) = 1/2 x n (Fe) = 1/2 x 0.12 = 0.06 mol

m (Fe₂O₃) = n x M = 0.06 x 159.7 = 9.6 g