Answer
(a) $S_8 + 8O_2 -- \gt 8SO_2$
(b) 20.0 $S_8$ molecules.
(c) 160. $SO_2$ molecules.
Work Step by Step
(a) - Balance the equation:
$S_8 + O_2 -- \gt SO_2$
- Balance $S$:
The subscript of the sulfur on the reactants is "8", so we should put that number as the coefficient of $SO_2$
$S_8 + O_2 -- \gt 8SO_2$
- Balance $O$:
There is a total of 16 oxygens in $8SO_2$, so we should put a "$\frac{16}{2}$" (which is 8), as the coefficient of $O_2$:
$S_8 + 8O_2 -- \gt 8SO_2$
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Use the coefficients as conversion factors:
(b) According to the balanced equation, the $O_2$ to $S_8$ ratio is 8:1
$160.O_2 \times \frac{1S_8}{8O_2} = 20.0$ $S_8$
(c) According to the same equation for each 8 oxygen molecules, 8 sulfur dioxides will be produced:
$160.O_2 \times \frac{8SO_2}{8O_2} = 160.$ $SO_2$
