Production of iron in blast furnace follows the following equation Fe 3 O 4( s )+4 CO ( g ) arrow 3 Fe ( l )+4 CO 2( g ) when 4.640 kg of Fe 3 O 4 and 2.520 kg of CO are allowed to react then the amount of iron (in g ) produced is: [Given: Molar Atomic mass ( g mol -1): Fe =56 Molar Atomic mass ( g mol -1): 0=16 Molar Atomic mass ( g mol -1):= C =12

Question

Production of iron in blast furnace follows the following equation Fe 3 O 4( s )+4 CO ( g ) arrow 3 Fe ( l )+4 CO 2( g ) when 4.640 kg of Fe 3 O 4 and 2.520 kg of CO are allowed to react then the amount of iron (in g ) produced is: [Given: Molar Atomic mass ( g mol -1): Fe =56 Molar Atomic mass ( g mol -1): 0=16 Molar Atomic mass ( g mol -1):= C =12 (A) 1400 (B) 2200 (C) 3360 (D) 4200

Tardigrade Admin · Accepted Answer

Moles of Fe 3 O 4=(4.640 × 103/232)=20 Moles of CO =(2.52 × 103/28)=90 So limiting Reagent = Fe 3 O 4 So moles of Fe formed =60 Weight of Fe =60 × 56=3360 gms

1400

2200

3360

4200

Moles of Fe3​O4​=2324.640×103​=20 Moles of CO=282.52×103​=90 So limiting Reagent =Fe3​O4​ So moles of Fe formed =60 Weight of Fe=60×56=3360gms

Moles of $F e_{3} O_{4} = \frac{4.640 \times 1 0 ^{3}}{232} = 20$
Moles of $CO = \frac{2.52 \times 1 0 ^{3}}{28} = 90$
So limiting Reagent $= F e_{3} O_{4}$
So moles of $F e$ formed $= 60$
Weight of $F e = 60 \times 56 = 3360 g m s$