Question

How many iron atoms would be produced by 4.5g of Fe2O3 reacting eith an excess of magnesium

Answer 1

Hope this helps you.

Answer 2

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 $0.34\times 10^{23}atoms$