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* 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}=(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
`(2)/(1)* 0.028=0.056` moles of
`Fe`

Now we have to calculate the mass of Fe.

`\text{Atoms of}Fe=\text{Moles of }Fe* \text{Avogadro's Number}`

`\text{Atoms of}Fe=(0.056mole)* (6.023* 10^(23))=0.34* 10^(23)atoms`

Therefore, the number of atoms will be
`0.34* 10^(23)atoms`