Answer : \(0.34\times 10^{23}atoms\)
Explanation : Given,
Mass of \(Fe_2O_3\) = 4.5 g
Molar mass of \(Fe_2O_3\) = 159.69 g/mole
Molar mass of Fe = 56 g/mole
First we have to calculate the moles of \(Fe_2O_3\).
\(\text{Moles of }Fe_2O_3=\frac{\text{Mass of }Fe_2O_3}{\text{Molar mass of }Fe_2O_3}=\frac{4.5g}{159.69g/mole}=0.028moles\)
Now we have to calculate the moles of Fe.
The balanced chemical reaction will be,
\(Fe_2O_3+3Mg\rightarrow 2Fe+3MgO\)
From the balanced reaction we conclude that
As, 1 moles of \(Fe_2O_3\) react to give 2 moles of \(Fe\)
So, 0.028 moles of \(Fe_2O_3\) react with \(\frac{2}{1}\times 0.028=0.056\) moles of \(Fe\)
Now we have to calculate the mass of Fe.
\(\text{Atoms of}Fe=\text{Moles of }Fe\times \text{Avogadro's Number}\)
\(\text{Atoms of}Fe=(0.056mole)\times (6.023\times 10^{23})=0.34\times 10^{23}atoms\)
Therefore, the number of atoms will be \(0.34\times 10^{23}atoms\)
