Question

To make `102*g` of alumina, how many grams of oxygen gas are necessary?

Answer 1

`1.26*mol` of dioxygen gas....

Explanation:

We need a stoichiometric equation to show mass and charge equivalence...

And so....

`underbrace(2Al(s))_(54.0*g) + underbrace(3/2O_2(g))_(48.0*g) rarr underbrace(Al_2O_3(g))_(102.0*g)`

Mass is balanced, and charge is balanced....and an `0.84*mol` quantity of aluminum was specified...

And so....`"mass of dioxygen"=0.84*molxx3/2*molxx32.00*g*mol^-1=40.3*g`