Question

Question #42d80

Answer 1

1.12 L

Explanation:

`S_8 + 8O_2 => 8SO_2`

=> 1 mole `S_8` => 8 moles `SO_2`

`"moles" S_8 = (1.6 g)/(256 g/"mole") = 0.00625 "mole"`

=> 0.00625 mole `S_8` => 0.00625(8) mole `SO_2` = 0.05 mole `S_8`

=> at STP => 0.05 mole `SO_2` => 0.05 mole ( 22.4 L/mole at STP ) = 1.12 Liters `SO_2`