Question

How much dioxygen is required to oxidize a `4*mol` quantity of aluminum metal?

Answer 1

If the reaction involved is this one

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

the answer is 11.1 moles of `O_2`

Explanation:

I'm going to assume that the reaction you are referring to is

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

In which case, the ratio of `("moles of "O_2 " required")/("moles of "Al " used") = 3/4`

Write an equation that has this ratio equal to a similar one in which you use the numbers given in this problem, with `x` representing the unknown amount of `O_2`:

`("moles of "O_2 " required")/("moles of "Al " used") = 3/4 = x/14.8`

Solve for `x`

`x=(3/4) (14.8) = 11.1 " moles of "O_2`

Answer 2

We need `11.1` moles of `O_2(g)`.

Explanation:

A stoichiometric equation is required:

`2Al(s) + 3/2O_2(g) rarr Al_2O_3(s)`

Given the equation, for each equiv metal `3/4` equiv dioxygen gas are required, i.e. `14.8*molxx3/4=11.1*"mol"` `O_2` gas.

Given standard condtions, how many litres of gas does this represent?