Question

What expression approximates the volume of O2 consumed, measure at STP, when 55 g of Al reacts completely with excess O2?2 Al(s) + 3 O2(g) ---> 2 Al2O3(s)0.5 x 1.5 x 22.40.5 x 0.67 x 22.42 x 0.67 x 22.42 x 1.5 x 22.4

Answer 1

Answer : The volume of
`O_2` consumed are,
`0.75* 2* 22.4` L.

Explanation :

The balanced chemical reaction will be:

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

First we have to calculate the moles of Al.

`\text{Moles of }Al=\frac{\text{Mass of }Al}{\text{Molar mass of }Al}`

Molar mass of Al = 27 g/mole

`\text{Moles of }Al=(55g)/(27g/mol)=2mole`

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

From the reaction we conclude that,

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

So, 2 mole of Al react with
`(3)/(4)* 2=0.75* 2` moles of
`O_2`

Now we have to calculate the volume of
`O_2` consumed.

As we know that, 1 mole of substance occupies 22.4 liter volume of gas.

As, 1 mole of
`O_2` occupies 22.4 liter volume of
`O_2` gas

So,
`0.75* 2` mole of
`O_2` occupies
`0.75* 2* 22.4` liter volume of
`O_2` gas

Therefore, the volume of
`O_2` consumed are,
`0.75* 2* 22.4` L.