Cl2g-2Br-excess-What are the final product of the above reaction equation-

The molar ionic conductance at infinite dilution of $M g^{2 +}$ and $C l^{-}$ ions are $106.1$ and $76.3 {S cm}^{2} {mol}^{- 1}$ respectively. The molar conductance of a solution of $M g C l_{2}$ at infinite dilution is:

⊝ 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

Q. Calculate

△ H_{r}^{o}

for the reaction

C H_{2} C l_{2} (g) \to C (g) + 2 H (g) + 2 C l (g) .

The average bond enthalpies of C - H and C - Cl bonds are

414 k J m o l^{- 1}

and

330 k J m o l^{- 1} .

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Cl2(g)+2Br−(excess)→What are the final product of the above reaction equation?

The correct option is C 2 mol of Cl− ions and 1 mol of Br2The given reaction may be expressed asCl2(g)+2Br−(excess)→2Cl−+Br2(g)Hence, 2 moles of Cl− and one mole of Br2 will be formed

$C l_{2} (g) + 2 B r^{-} (e x c e s s) \to$
What are the final product of the above reaction equation?

The correct option is C $2$ mol of $C l^{-}$ ions and $1$ mol of $B r_{2}$
The given reaction may be expressed as

$C l_{2} (g) + 2 B r^{-} (e x c e s s) \to 2 C l^{-} + B r_{2} (g)$

Hence, 2 moles of $C l^{-}$ and one mole of $B r_{2}$ will be formed