The mass of iron converted into Fe3O4 by the action of 18 g of steam is : (Given : Molar mass of H, O and Fe are 1, 16 and 56 g mol-1 respectively) Assume iron is present in excess :
- 2.1 g
- 4.2 g
- 21 g
- 42 g
Answer
(4) 42 g
Solution
Reaction of steam with red-hot iron:
3Fe(s) + 4H2O(g) Fe3O4(s) + 4H2(g)
∵ Moles of steam = 18 g18 g mol-1 = 1 mol
From the equation, 4 mol H2O react with 3 mol Fe
⇒ moles of Fe converted = 34 1 = 0.75 mol
∴ Mass of Fe = 0.75 56 = 42 g
Answer: (4)