In the following redox reaction ${C r}_{2} O_{7}^{2 -} + {F e}^{2 +} ⟶ {F e}^{3 +} + {C r}^{3 +}$ 1 mol of $C r_{2} O_{7}^{2 -}$ oxidizes x mol of ${F e}^{+ 2}$ . Find x.

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Solution

${C r}_{2} O_{7}^{2 -} + {F e}^{2 +} \to {F e}^{3 +} + {C r}^{3 +}$
After balancing the given reaction we will get following balanced chemical reaction:
${C r}_{2} O_{7}^{2 -} + 14 H^{+} + 3 {F e}^{2 +} \to {2 C r}^{3 +} + 3 {F e}^{3 +} + 7 H_{2} O$
$∴$ $1$ mole of ${C r}_{2} O_{7}^{2 -}$ oxidize $3$ moles of ${F e}^{2 +}$ .
$∴$ $x = 3$

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