How many moles of Al_2O_3 can be prepared from 11*mol of aluminum?

1 Answer
anor277
May 24, 2017

5.5*molxx101.96*g*mol^-1=560.78*g...........

Explanation:

If I have 18 eggs, then how many dozen eggs do I have? I think you would quickly be able to respond that you have 3/2 "dozen", because "1 dozen" = 12.

If I have a mole of stuff, aluminum atoms, oxygen atoms, I have "Avogadro's number" of individual atoms, where "Avogadro's number"-=6.022xx10^23*mol^-1.

Here we have 11*mol of aluminum metal; given the stoichiometry we can make 5.5*mol Al_2O_3, which has a molar mass of 101.96*g*mol^-1.

And thus we take the product, "molar quantity"xx"molar mass" to get an actual mass.

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