Answer:
[tex]m_{Fe}=171.7gFe[/tex]
Explanation:
Hello!
In this case, since we have 245 of iron (III) oxide, we first need to compute the moles contained there:
[tex]n_{Fe_2O_3}=245gFe_2O_3*\frac{1molFe_2O_3}{159.7gFe_2O_3} =1.54molFe_2O_3[/tex]
Now, as 1 mole of iron (III) oxide is related to 2 moles of iron, due to iron's subscript in the molecule, we get the moles of iron itself:
[tex]n_{Fe}=1.54molFe_2O_3*\frac{2molFe}{1molFe_2O_3} =3.08molFe[/tex]
And the mass is computed based on the atomic mass of iron:
[tex]m_{Fe}=3.08molFe*\frac{55.8gFe}{1molFe} \\\\m_{Fe}=171.7gFe[/tex]
Best regards!
