You visited us 1 times! Enjoying our articles? Unlock Full Access!

Equivalent Mass

The number of...

Question

The number of moles of $K_{2} {C r}_{2} O_{7}$ reduced by $3$ mol of ${S n}^{2 +}$ ions is :

Open in App

Solution

The redox changes are shown below:
$6 e^{-} + {C r}_{2} O_{7}^{2 -} ⟶ 2 {C r}^{3 +}$
${S n}^{2 +} ⟶ {S n}^{4 +} + 2 e^{-}$
Equivalent of ${C r}_{2} O_{7}^{2 -} =$ Equivalent of ${S n}^{2 +}$
$\Rightarrow 1$ Eq. $\equiv 1$ Eq.

$\frac{1}{6}$ mol $\equiv \frac{1}{2}$ mol

$1$ mol of ${C r}_{2} O_{7}^{2 -} = 3$ mol of ${S n}^{2 +}$

Suggest Corrections

Similar questions

K_{2} {C r}_{2} O_{7}

1

{S n}^{2 +}

K M n O_{4}

K_{2} {C r}_{2} O_{7}

{S n}^{2 +}

{S n}^{+ 4}

({C r}_{2} O_{7}^{- 2})

{C r}^{+ 3}

K_{2} {C r}_{2} O_{7}

K_{2} C r_{2} O_{7}

S n^{2 +}

K_{2} C r_{2} O_{7}

S n^{2 +}

Join BYJU'S Learning Program