You need 158.70 grams of Fe2O3 to produce 111 grams of Fe. This is calculated by using the molar masses and stoichiometric relationship of the two compounds.
Solution:
MM Fe = 55.845 g/mol
MM Fe2O3 = 159.69 g/mol
Fe: Fe2O3 = 2 mol:1 mol
11 g FE (1 mol Fe/55.845 g Fe) (1 mol Fe2O3/2 mol Fe) (159.69 g Fe2O3 / 1 mole Fe2O3) = 158.70 grams Fe2O3
Answer:
The flocculation basin often has a number of compartments with decreasing mixing speeds as the water advances through the basin. ... This compartmentalized chamber allows increasingly larger floes to form without being broken apart by the mixing blades.
