Question

56 g of red hot iron is treated with 36 ml water the wight of fe3o4 produced would be [fe=56]

Answer 1

Answer: 76.4 g of Fe3O4 produced from the reaction of 56 g red hot iron with 36 mL of water.

Explanation:Given that,

Amount of red hot iron = 56 g

Moles of red hot iron in 56 g = 56 g/56 g/mol = 1 mol

Volume of water reacted with red hot iron = 36 mL

Mass of water reacted with red hot iron = 36g ( density of water is 1 g/mL at room temperature)

Moles of water in 36 g = 36 g/18 g/mol = 2 mol

General Reaction of red hot iron with water:

3Fe + 4H2O ----> Fe3O4 + 4H2

Theoretically,

3 mol of Iron reacts with enough water to produce = 1 mol of Fe3O4

1 mol of Iron reacts with enough water to produce = (1/3)×1 mol of Fe3O4

= 0.33 mol of Fe3O4

Molar mass of Fe3O4 = 231.533 g/mol

Weight of 0.33 mol of Fe3O4 = 0.33 mol ×231.533 g/mol

= 76.4 g