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

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Answer 1 · 2018-02-16T16:54:26

$1.26*mol$ of dioxygen gas....

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

To make 102*g102*g of alumina, how many grams of oxygen gas are necessary?