[tex]8H₂S_{(g)}+4O_{2(g)\text{ }}→\text{ }8H₂O_{(g)}+S_{8(g)}[/tex][tex]v=\frac{-1}{8}\frac{\Delta\lbrack H_2S\rbrack}{\Delta t}=\frac{-1}{4}\frac{\Delta\lbrack O_2\rbrack}{\Delta t}=\frac{1}{8}\frac{\Delta\lbrack H_2O\rbrack}{\Delta t}=\frac{1}{1}\frac{\Delta\lbrack S_8\rbrack}{\Delta t}[/tex][tex]\frac{-1}{8}\frac{\Delta\lbrack H_2S\rbrack}{\Delta t}=\frac{1}{1}\frac{\Delta\lbrack S_8\rbrack}{\Delta t}[/tex][tex]\frac{\Delta\lbrack S_8\rbrack}{\Delta t}=\text{ }\frac{-1}{8}\frac{\Delta\lbrack H_2S\rbrack}{\Delta t}=\frac{-1}{8}(-0.021\text{ M/s\rparen = 0.0026 M/s}[/tex][tex]v=\frac{\Delta\lbrack S_8\rbrack}{\Delta t}=\text{ 0.0026 M/s}[/tex]
Part C the answer is Δ[S8]/Δt = 0.0026 M/s
Parte D the answer is v = 0.0026 M/s
