Chemistry - Redox Reactions Question with Solution | TestHub

In acid, ${\mathrm{K}}_2\mathrm{C}{\mathrm{r}}_2{\mathrm{O}}_7$ has $n$ = 6, so 14.7/294 = 0.05 mol = 0.3 g-eq; we need exactly 0.3 g-eq of reductant. (a) $\mathrm{HCl}$: 10.95/36.5 = 0.3 mol, $n$ = 1 $\to$ 0.3 g-eq. (b) $\mathrm{FeS}{\mathrm{O}}_4$: 38/152 = 0.25 mol, $n$ = 1 $\to$ 0.25 g-eq (insufficient). (c) ${\mathrm{H}}_2{\mathrm{C}}_2{\mathrm{O}}_4$: 13.5/90 = 0.15 mol, $n$ = 2 $\to$ 0.3 g-eq. (d) $\mathrm{S}{\mathrm{O}}_2$: 32/64 = 0.5 mol, $n$ = 2 $\to$ 1.0 g-eq (excess). Hence (a), (c).