Question

Aluminum oxide (used as an adsorbent or a catalyst for organic reactions) forms when aluminum reacts with oxygen. 4al(s) + 3o2(g) ® 2al2o3(s) [balanced] a mixture of 82.49 g of aluminum ( = 26.98 g/mol) and 117.65 g of oxygen ( = 32.00 g/mol) is allowed to react. what mass of aluminum oxide ( = 101.96 g/mol) can be formed?

Answer 1

Aluminum oxide (used as an adsorbent or a catalyst for organic reactions) forms when aluminum reacts with oxygen. 4al(s) + 3o2(g) ® 2al2o3(s) [balanced] a mixture of 82.49 g of aluminum ( = 26.98 g/mol) and 117.65 g of oxygen ( = 32.00 g/mol) is allowed to react. what mass of aluminum oxide ( = 101.96 g/mol) can be formed?

Answer 2

Answer : The mass of aluminium oxide formed can be 155.8 grams.

Solution : Given,

Mass of Al = 82.49 g

Mass of
`O_2` = 117.65 g

Molar mass of Al = 26.98 g/mole

Molar mass of
`O_2` = 32.00 g/mole

Molar mass of
`Al_2O_3` = 101.96 g/mole

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

`\text{ Moles of }Al=\frac{\text{ Mass of }Al}{\text{ Molar mass of }Al}=(82.49g)/(26.98g/mole)=3.057moles`

`\text{ Moles of }O_2=\frac{\text{ Mass of }O_2}{\text{ Molar mass of }O_2}=(117.65g)/(32.00g/mole)=3.677moles`

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

`4Al(s)+3O_2(g)\rightarrow 2Al_2O_3(s)`

From the balanced reaction we conclude that

As, 4 mole of
`Al` react with 3 mole of
`O_2`

So, 3.057 moles of
`Al` react with
`(3.057)/(4)* 3=2.293` moles of
`O_2`

From this we conclude that,
`O_2` is an excess reagent because the given moles are greater than the required moles and
`Al` is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of
`Al_2O_3`

From the reaction, we conclude that

As, 4 mole of
`Al` react to give 2 mole of
`Al_2O_3`

So, 3.057 mole of
`Al` react to give
`3.057* (2)/(4)=1.528` moles of
`Al_2O_3`

Now we have to calculate the mass of
`Al_2O_3`

`\text{ Mass of }Al_2O_3=\text{ Moles of }Al_2O_3* \text{ Molar mass of }Al_2O_3`

`\text{ Mass of }Al_2O_3=(1.528moles)* (101.96g/mole)=155.8g`

Therefore, the mass of aluminium oxide formed can be 155.8 grams.