When Fe 0.93 O is heated in presence of oxygen, it converts to Fe 2 O 3. The number of correct statement/s from the following is A. The equivalent weight of Fe 0.93 O is Molecular wejght 0.79 B. The number of moles of Fe 2+ and Fe 3+ in 1 mole of Fe 0.93 O is 0.79 and 0.14 respectively C. Fe 0.93 O is metal deficient with lattice comprising of cubic closed packed arrangement of O 2- ions D. The % composition of Fe 2+ and Fe 3+ in Fe 0.93 O is 85 % and 15 % respectively

Question

When Fe 0.93 O is heated in presence of oxygen, it converts to Fe 2 O 3. The number of correct statement/s from the following is A. The equivalent weight of Fe 0.93 O is Molecular wejght 0.79 B. The number of moles of Fe 2+ and Fe 3+ in 1 mole of Fe 0.93 O is 0.79 and 0.14 respectively C. Fe 0.93 O is metal deficient with lattice comprising of cubic closed packed arrangement of O 2- ions D. The % composition of Fe 2+ and Fe 3+ in Fe 0.93 O is 85 % and 15 % respectively (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

A: Fe 0.93 O arrow Fe 2 O 3 nf =(3-(200/93)) × 0.93 nf =0.79 B: 2 x+(0.93-x) × 3=2 x=0.79 Fe 2+=0.79, Fe 3+=0.21 C: Fact D: % Fe 2+=(0.79/0.93) × 100=85 % ; Fe 3+=15 %

A : Fe0.93​O→Fe2​O3​ nf=(3−93200​)×0.93 nf=0.79 B: 2x+(0.93−x)×3=2 x=0.79 Fe2+=0.79,Fe3+=0.21 C : Fact D : %Fe2+=0.930.79​×100=85%;Fe3+=15%

A : $F e_{0.93} O \to F e_{2} O_{3}$
$n f = (3 - \frac{200}{93}) \times 0.93$
$n f = 0.79$
B: $2 x + (0.93 - x) \times 3 = 2$
$x = 0.79$
$F e^{2 +} = 0.79, F e^{3 +} = 0.21$
C : Fact
D : $% F e^{2 +} = \frac{0.79}{0.93} \times 100 = 85%; F e^{3 +} = 15%$