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How many grams of #Al_2O_3# would form if 12.5 moles of #Al# burned?

Answer 1

Wyatt Adkins

Over #600# #g# of aluminum oxide would result.

For the combustion of aluminum, the stoichiometric equation is:

#2Al(s) + 3/2O_2(g) rarr Al_2O_3(s)#

If #12.5# #mol# of aluminum were consumed, clearly #6.25# #mol# aluminum oxide would result.

#"Mass of aluminum oxide"# #=# #6.25*molxx101.96*g*mol^-1# #=# #??# #g#.

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Answer 2

Jose Ayala

To find the mass of Al2O3 formed, you need to use the balanced chemical equation for the combustion of aluminum:

4Al + 3O2 -> 2Al2O3

From the equation, you can see that 4 moles of Al react to form 2 moles of Al2O3.

Using the given number of moles of Al (12.5 moles), you can set up a proportion:

(12.5 moles Al) / (4 moles Al) = (x moles Al2O3) / (2 moles Al2O3)

Solving for x gives:

x = (12.5 moles Al) * (2 moles Al2O3) / (4 moles Al) = 6.25 moles Al2O3

Now, you can use the molar mass of Al2O3 (101.96 g/mol) to find the mass:

Mass = (6.25 moles Al2O3) * (101.96 g/mol Al2O3) = 637.25 g Al2O3

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.