(i) K 4 Fe(CN) 6 + 3H 2 SO 4 
2K 2 SO 4 + FeSO 4 + 6HCN
(ii) 6HCN + 12H 2 O 
6HCOOH + 6NH 3
(iii)
3(NH 4 ) 2 SO 4 LR = H 2 SO 4
6CO + 6H 2 O Above steps of reactions occur in a container starting with one mole of K 4 [Fe(CN) 6 ], 5 mole of H 2 SO 4 and enough water. Find out the limiting reagent in step (i) and calculate maximum moles of CO gas and (NH 4 ) 2 SO 4 that can be produced. LR = K 4 Fe(CN) 6 ,
Text Solution
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(b,c)
(i) K 4 Fe(CN) 6 + 3H 2 SO 4 
2K 2 SO 4 + FeSO 4 + 6HCN
1 mole 5 mole
Limiting 1/1 5/3
reagent
(1–1) (5–3 × 1) 2 × 1 1 × 1 6 × 1
0 mole 2 mole 2 mole 1 mole 6 mole
Limiting reagent in step (i) is K 4 [Fe(CN) 6 ]
(ii) 6HCN + 12H 2 O 
6HCOOH + 6NH 3
6 mole (excess) 0 0
0 6 mole 6 mole
(iii) 6NH 3 + 3H 2 SO 4 
3(NH 4 ) 2 SO 4
6 mole 2 mole 0
Limiting 6/6 2/3
reagent
(6– 
× 6) (2– 
× 3) (3 × 
)
2 mole 0 mole 2 mole
6HCOOH 
6CO + 6H 2 O
6 mole 0 mole 0 mole
0 mole 6 mole 6 mole
Limiting reagent in step (i) is K 4 [Fe(CN) 6 ]
(NH 4 ) 2 SO 4 = 2 mol
CO gas = 6 mol
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