Respuesta :
Answer: The mass of iron (II) oxide that must be used in the reaction is 30.37
Explanation:
The given chemical reaction follows:
[tex]6FeO(s)+O_2(g)\rightarrow 2Fe_3O_4(s);\Delta H^o=-635kJ[/tex]
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, [tex]\frac{6}{635}\times 44.7=0.423mol[/tex] of iron (II) oxide is reacted.
Now, calculating the mass of iron (II) oxide by using the equation:
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex]
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:
[tex]0.423mol=\frac{\text{Mass of FeO}}{71.8g/mol}\\\\\text{Mass of FeO}=(0.423mol\times 71.8g/mol)=30.37g[/tex]
Hence, the mass of iron (II) oxide that must be used in the reaction is 30.37
