How many liters of H2 at STP are required to react with 2.3 g of Fe3O4 to give FeO and H2O?

0.22L of H2

Explanation:

We'll begin by writing the balanced equation for the reaction. This is given below:

Fe3O4 + H2 - > 3FeO + H2O

Next, we shall determine the number of mole present in 2.3g of Fe3O4. This is illustrated below:

Molar mass of Fe3O4 = (56x3) + (16x4) = 232g/mol

Mass of Fe3O4 = 2.3g

Mole of Fe3O4 = ... ?

Mole = Mass/Molar Mass

Mole of Fe3O4 = 2.3/232 = 9.91x10^-3 mole.

Next, we shall determine the number of mole H2 needed for the reaction. This is illustrated below:

From the balanced equation above,

1 mole of Fe3O4 reacted with 1 mole of H2.

Therefore, 9.91x10^-3 mole of Fe2O3 will also react with 9.91x10^-3 mole of H2.

Finally, we can obtain the volume of H2 required for the reaction as follow:

1 mole of a gas occupy 22.4L at stp.

Therefore, 9.91x10^-3 mole of H2 will occupy = 9.91x10^-3 x 22.4 = 0.22L

Therefore, 0.22L of H2 is required for the reaction.