Find the Equivalent weight of ${A s}_{2} O_{3}$ in the following equation:
${A s}_{2} O_{3} + {2 I}_{2} + {2 H}_{2} O \to {A s}_{2} O_{5} + 4 H I$

Open in App

Solution

$A s_{2} O_{3} + 2 I_{2} + 2 H_{2} O ⟶ A s_{2} O_{5} + 4 H I$
Oxidation state of $A s$ in $A s_{2} O_{5} = + 3$
Oxidation state of $A s$ in $A s_{2} O_{5} = + 5$
$∴$ Change in oxidation state of $A s = 2$
$∴$ Change in oxidation state of $A s_{2} O_{3} = 4$
$∴$ Equivalent weight of $A s_{2} S_{3} = \frac{M o l e c u l a r w e i g h t o f A s_{2} O_{3}}{4}$
= $\frac{M}{4} = \frac{246}{4}$
= $61.5 g$

Suggest Corrections

Similar questions

{A s}_{2} O_{3}

{A s}_{2} O_{3}

H_{3} A s O_{4}

K M n O_{4}

0.02 M

K M n O_{4}

1

{A s}_{2} O_{3}

{N a}_{2} S_{2} O_{3}

2 {N a}_{2} S_{2} O_{3} + I_{2} ⟶ {N a}_{2} S_{4} O_{6} + 2 N a I

\begin{matrix} Column -I & Column -II a . N_{2} O & (i) Basic b . B a O & (ii)Amphoteric c . A s_{2} O_{3} & (iii) Acidic d . C l_{2} O_{7} & (iv) Neutral \end{matrix}

A s_{2} S_{3}

M

{A s}_{2} S_{3} + H^{+} + {N O}_{3}^{-} ⟶ N O + H_{2} O + A s O_{4}^{3 -} + S O_{4}^{2 -}

Join BYJU'S Learning Program