-OH - S2O2-3- Br2- SO2-4 - Br- In the above reaction- starting with 0-15 mol of Br2- 0-01 mol of S2O2-3 and 0-4 mol of -OH ions- x moles of -OH ions are left in the solution after the reaction is complete- Find the value of 10x-

Balance the equation in basic medium nbsp; nbsp; nbsp; nbsp; nbsp; nbsp; nbsp; nbsp; nbsp; 10OH+S_2O^2 _3→ 2SO^2 _4+5H_2O+8e^ nbs ...

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Standard XII

Chemistry

Limiting Reagent or Reactant

⊝OH + S2O2-3+...

Question

$⊝ O H + S_{2} O_{3}^{2 -} + B r_{2} \to S O_{4}^{2 -} + B r^{⊝}$
In the above reaction, starting with $0.15$ mol of $B r_{2}, 0.01$ mol of $S_{2} O_{3}^{2 -}$ and $0.4$ mol of $⊝ O H$ ions, $x$ moles of $⊝ O H$ ions are left in the solution after the reaction is complete. Find the value of $10 x$ .

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Solution

Balance the equation in basic medium
$10 O H + S_{2} O_{3}^{2 -} \to 2 S O_{4}^{2 -} + 5 H_{2} O + 8 e^{-}$
$2 e^{-} + B r_{2} \to 2 B r^{⊝}] \times 4$
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$4 B r_{2} + S_{2} O_{3}^{2 -} + 10 ⊝ O H \to 2 S O_{4}^{2 -} + 8 B r^{⊝} + 5 H_{2} O$
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$⎡ ⎢ ⎣ \begin{matrix} M o l e s b e f o r e r e a c t i o n \end{matrix} ⎤ ⎥ ⎦$ $(0.15)$ $(0.01)$ $(0.4)$ $(0)$ $(0)$ $(0)$
$⎡ ⎢ ⎢ ⎢ ⎣ \begin{matrix} M o l e s r e a c t e d a n d f o r m e d \end{matrix} ⎤ ⎥ ⎥ ⎥ ⎦$ $(0.01 \times 4)$ $(0.01)$ $(0.01 \times 10)$ $(0.01 \times 2)$ $(0.01 \times 8)$ $(0.01 \times 5)$
$[\begin{matrix} M o l e s l e f t \end{matrix}]$ $(0.15 - 0.04) (= 0.11)$ $(0.4 - 0.1) (= 0.3)$ $(0.02)$ $(0.08)$ $(0.05)$
So, molar ratio of combination is:
$B r_{2} : S_{2} O_{3}^{2 -} : ⊝ O H :: 4 : 1 : 10$
$[⊝ O H]$ left after reaction $= 0.3$ mol

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⊝OH+S2O2−3+Br2→SO2−4+Br⊝ In the above reaction, starting with 0.15 mol of Br2, 0.01 mol of S2O2−3 and 0.4 mol of ⊝OH ions, x moles of ⊝OH ions are left in the solution after the reaction is complete. Find the value of 10x.