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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