Question

Given the following equation: 4 Fe + 3022 Fe2O3 What mass of iron is needed

to react with 12 grams of oxygen gas? *​

Answer 1

Answer : The mass of iron needed is 29.5 grams.

Explanation : Given,

Mass of
`O_2` = 12 g

Molar mass of
`O_2` = 32 g/mol

First we have to calculate the moles of
`O_2`.

`\text{Moles of }O_2=\frac{\text{Given mass }O_2}{\text{Molar mass }O_2}`

`\text{Moles of }O_2=(12g)/(32g/mol)=0.375mol`

Now we have to calculate the moles of
`Fe`

The balanced chemical equation is:

`4Fe+3O_2\rightarrow 2Fe_2O_3`

From the reaction, we conclude that

As, 3 moles of
`O_2` react with 4 moles of
`Fe`

So, 0.375 mole of
`O_2` react with
`(4)/(3)* 0.375=0.5` mole of
`Fe`

Now we have to calculate the mass of
`Fe`

`\text{ Mass of }Fe=\text{ Moles of }Fe* \text{ Molar mass of }Fe`

Molar mass of
`Fe` = 59 g/mole

`\text{ Mass of }Fe=(0.5moles)* (59g/mole)=29.5g`

Therefore, the mass of iron needed is 29.5 grams.