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

1 Answer
Jun 9, 2018

Answer:

Approx. #6.5*g# of sulfur….

Explanation:

We interrogate the stoichiometric reaction....

#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#...

And thus we need #"3 equivs"# of sulfur oxidant...

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