Aluminum metal will combine with oxygen gas to produce aluminum oxide?

if 1.3 g of aluminum is placed into a container containing 0.96 g of oxygen gas, what mass of aluminum oxide will form? What is the equation?

The balanced chemical equation is : 4 Al + 3 02 => 2 Al2O3

You have to first find the limiting reactant (reactant that is all used up in the reaction).
To find limiting reactant, find the number of moles of each reactant and multiply by the molar ratio according to the product. The lower number will be your limiting reactant.

nAl=m/M
nAl=(1.3g)/(26.98)
nAl=0.048 ========> (2/4) ===========> nAl2O3=0.024

nO2=m/M
nO2=(0.96g)/(32)
nO2=0.030 ========> (2/3) ===========> nAl2O3= 0.020

This concludes that the oxygen gas will be all used up, and will be the limiting reactant.
Now to find the number of moles of Aluminum Oxide (which we already did), take the moles of the limiting reactant and multiply by the molar ratio (0.020 mol).

SO,

mAl2O3= Mn
mAl2O3= (101.96)(0.020)
mAl2O3= 2.04g

The mass of the product will be 2.04g.

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