Please calculate the atom economy and the estimated e-factor (assuming 100% yield))

Question

Please calculate the atom economy and the estimated e-factor (assuming 100% yield))

Reaction: Fe2+(aq) + H2CC204(aq) + 2H20(1) → FEC204(s) + 2 H30+(aq) 2 FEC204(s) + H2O2(aq) + 2 H+(aq) + C2042-(aq) → Fe2(C2O4)3(s) + 4 H2O(I) Fe2(C204)3 (s) + 3 H2C204 (aq) + 6 H2O(1) → 2 [Fe(C204)3]3 -(aq) + 6 H3O+ (aq) 3 K+(aq) + [Fe(C204)3]3-(aq) → K3[Fe(C2O4)3)(s) 



Answer 1

Fe²⁺(aq) + H₂CC₂O₄(aq) + 2H₂0(l) ==> FeC₂O₄(s) + 2H₃O⁺(aq)

Atom economy is essentially the % of desired product relative to total product (on a mass basis), or ...

molar mass of desired product / sum of molar mass of all products (x100%)

molar mass FeC₂O₄ = 144 g/mol

molar mass H₃O⁺ = 19 g/mol

Sum = 144 + 19 = 163

Atom economy = 144 / 163 (x100%) = 88.3%

e factor = mass of waste product / mass of desired product

e factor = 19 / 144 = 0.132

Do the others the same way