The weight of iron which will be converted into its oxide (Fe3O4) by the action of 18 g of steam on it will be (at. wt. of Fe = 56)

Question

The weight of iron which will be converted into its oxide (Fe3O4) by the action of 18 g of steam on it will be (at. wt. of Fe = 56) (A) 168 g (B) 84 g (C) 42 g (D) 21 g

Tardigrade Admin · Accepted Answer

3Fe+4H2O xrightarrowFe3O4+4H2 3× 56g 4× 18g 4× 18g H2O≡ 3× 56g Fe ∴ 18g H2O=(3× 56/4)=42g

Q. The weight of iron which will be converted into its oxide $(F e_{3} O_{4})$ by the action of 18 g of steam on it will be (at. wt. of Fe = 56)

168 g

84 g

42 g

21 g

$3 F e + 4 H_{2} O F e_{3} O_{4} + 4 H_{2}$ $3 \times 56 g 4 \times 18 g$ $4 \times 18 g H_{2} O \equiv 3 \times 56 g F e$ $∴$ $18 g H_{2} O = \frac{3 \times 56}{4} = 42 g$

Q. The weight of iron which will be converted into its oxide (Fe3​O4​) by the action of 18 g of steam on it will be (at. wt. of Fe = 56)