Question

What mass of iron(II) oxide must be used in the reaction given by the equation below to release 44.7 kJ? 6FeO(s) + O2(g) => 2Fe3O4(s) ΔH° = -635 kJ Calculate your answer in g. Enter it with two decimal places and no units.

Answer 1

Answer: The mass of iron (II) oxide that must be used in the reaction is 30.37

Explanation:

The given chemical reaction follows:

$6FeO(s)+O_2(g)\rightarrow 2Fe_3O_4(s);\Delta H^o=-635kJ$

By Stoichiometry of the reaction:

When 635 kJ of energy is released, 6 moles of iron (II) oxide is reacted.

So, when 44.7 kJ of energy is released, $\frac{6}{635}\times 44.7=0.423mol$ of iron (II) oxide is reacted.

Now, calculating the mass of iron (II) oxide by using the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Moles of iron (II) oxide = 0.423 moles

Molar mass of iron (II) oxide = 71.8 g/mol

Putting values in above equation, we get:

$0.423mol=\frac{\text{Mass of FeO}}{71.8g/mol}\\\\\text{Mass of FeO}=(0.423mol\times 71.8g/mol)=30.37g$

Hence, the mass of iron (II) oxide that must be used in the reaction is 30.37