2d(g) + 3e(g) + f(g) -------> 2g(g) + h(g)
As it can be seen from the equation that for every 3 moles of e consumed,=2 moles of g formed.
for every 1 moles of e consumed =[tex]\frac{2}{3}[/tex] moles of g formed
So, if e is decreasing at a rate of 0.16 mol s⁻¹
[tex]=\frac{2}{3}\times0.16 = 0.106 mol s^{-1}[/tex]
So when e is decreasing at 0.16 mol s⁻¹, g is increasing at a rate of 0.106 mol s⁻¹.
