## Chemistry 10th Edition

(a) $S_8 + 8O_2 -- \gt 8SO_2$ (b) 20.0 $S_8$ molecules. (c) 160. $SO_2$ molecules.
(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 $O_2$: $S_8 + 8O_2 -- \gt 8SO_2$ ----------------- 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: $160.O_2 \times \frac{8SO_2}{8O_2} = 160.$ $SO_2$