Question #42d80

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Answer 1 · 2017-06-20T21:54:48

1.12 L

$S_{8} + 8 O_{2} \Rightarrow 8 S O_{2}$ $S_8 + 8O_2 => 8SO_2$

=> 1 mole $S_{8}$ $S_8$ => 8 moles $S O_{2}$ $SO_2$

$moles S_{8} = \frac{1.6 g}{256 \frac{g}{mole}} = 0.00625 mole$ $"moles" S_8 = (1.6 g)/(256 g/"mole") = 0.00625 "mole"$

=> 0.00625 mole $S_{8}$ $S_8$ => 0.00625(8) mole $S O_{2}$ $SO_2$ = 0.05 mole $S_{8}$ $S_8$

=> at STP => 0.05 mole $S O_{2}$ $SO_2$ => 0.05 mole ( 22.4 L/mole at STP ) = 1.12 Liters $S O_{2}$ $SO_2$