If 10.0 g of aluminum sulfide are produced by the reaction of aluminum and sulfur, how many grams or sulfur were needed?

Answer 1

Approx. #6.5*g# of sulfur….

We examine the reaction that is stoichiometric.

#2Al(s) + 3S(s) stackrel(Delta)rarrAl_2S_3(s)#
Aluminum is oxidized, and sulfur is reduced (note that sometimes, we represent elemental sulfur, i.e. #"flowers of sulfur"#, as #S_8#..)
#"Moles of aluminum sulfide"=(10.0*g)/(150.16*g*mol^-1)=0.0666*mol#...

Thus, "3 equivs" of sulfur oxidant are needed.

#-=3xx0.0666*molxx32.06*g*mol^-1=??*g#
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Answer 2

Sulfur needed = 18.0 g

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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.

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.

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.

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.

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