How many #O_2# molecules are needed to react with 7.87 g of S?

1 Answer
anor277
Aug 18, 2016

Wee assume formation of the #S(IV)# oxide, #SO_2(g)#

Explanation:

#S + O_2(g) rarr SO_2(g)#

Moles of #S# #=# #(7.87*g)/(32.06*g*mol^-1)# #=# #0.245*mol#

And thus, by reaction stoichiometry, we need #0.245*mol# dioxygen gas.

So we need, #0.245*molxx6.022xx10^23*mol^-1" oxygen molecules"#.

How many oxygen atoms does this represent?

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