Answer : The mass of iron needed is 29.5 grams.
Explanation : Given,
Mass of [tex]O_2[/tex] = 12 g
Molar mass of [tex]O_2[/tex] = 32 g/mol
First we have to calculate the moles of [tex]O_2[/tex].
[tex]\text{Moles of }O_2=\frac{\text{Given mass }O_2}{\text{Molar mass }O_2}[/tex]
[tex]\text{Moles of }O_2=\frac{12g}{32g/mol}=0.375mol[/tex]
Now we have to calculate the moles of [tex]Fe[/tex]
The balanced chemical equation is:
[tex]4Fe+3O_2\rightarrow 2Fe_2O_3[/tex]
From the reaction, we conclude that
As, 3 moles of [tex]O_2[/tex] react with 4 moles of [tex]Fe[/tex]
So, 0.375 mole of [tex]O_2[/tex] react with [tex]\frac{4}{3}\times 0.375=0.5[/tex] mole of [tex]Fe[/tex]
Now we have to calculate the mass of [tex]Fe[/tex]
[tex]\text{ Mass of }Fe=\text{ Moles of }Fe\times \text{ Molar mass of }Fe[/tex]
Molar mass of [tex]Fe[/tex] = 59 g/mole
[tex]\text{ Mass of }Fe=(0.5moles)\times (59g/mole)=29.5g[/tex]
Therefore, the mass of iron needed is 29.5 grams.
