Cl 2 g -X - O H- YCl O 3 - -Z H 2 O-10 e -The coefficient X- Y and Z are-

The correct option is A 12, 2, 6 Cl_2+OH^ → ClO_3^ +zH_2O Cl_2→ 2ClO_3^ +12H^++10e^ +6H_2O Adding 12OH^ on both sides. Cl_2+6H_2O+12OH^ → 2 ...

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Byju's Answer

Standard XII

Chemistry

Stoichiometry

Cl 2 g +X - O...

Question

${C l}_{2} (g) + X - O H \to Y C l {O_{3}}^{-} + Z H_{2} O + 10 e^{-}$
The coefficient X, Y and Z are:

6, 2, 2

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5, 1, 3

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12, 2, 6

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12, 1, 6

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Solution

The correct option is A 12, 2, 6
$C l_{2} + O H^{-} \to C l O_{3}^{-} + z H_{2} O$
$C l_{2} \to 2 C l O_{3}^{-} + 12 H^{+} + 10 e^{-} + 6 H_{2} O$
Adding $12 O H^{-}$ on both sides.
$C l_{2} + 6 H_{2} O + 12 O H^{-} \to 2 C l O_{3}^{-} (12 H^{-} + 12 O H^{-}) + 10 e^{-}$
$\Rightarrow C l_{2} + 6 H_{2} O + 12 O H^{-} \to 2 C l O_{3}^{-} + 12 H_{2} O + 10 e^{-}$
$C L_{2} + 12 O H^{-} \to 2 C l O_{3}^{-} + 6 H_{2} O + 10 e^{-}$
$x = 12$ $y = 2$ $z = 6$

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Similar questions

C l_{2} (g) + X O H^{-} \to Y C l O_{3}^{-} + Z H_{2} O + 10 e^{-}

The coefficient X,Y and Z are:

Q. Find the angle between the following pairs of lines:
(i)

\frac{x + 4}{3} = \frac{y - 1}{5} = \frac{z + 3}{4} and \frac{x + 1}{1} = \frac{y - 4}{1} = \frac{z - 5}{2}

(ii)

\frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z - 3}{- 3} and \frac{x + 3}{- 1} = \frac{y - 5}{8} = \frac{z - 1}{4}

(iii)

\frac{5 - x}{- 2} = \frac{y + 3}{1} = \frac{1 - z}{3} and \frac{x}{3} = \frac{1 - y}{- 2} = \frac{z + 5}{- 1}

(iv)

\frac{x - 2}{3} = \frac{y + 3}{- 2}, z = 5 and \frac{x + 1}{1} = \frac{2 y - 3}{3} = \frac{z - 5}{2}

(v)

\frac{x - 5}{1} = \frac{2 y + 6}{- 2} = \frac{z - 3}{1} and \frac{x - 2}{3} = \frac{y + 1}{4} = \frac{z - 6}{5}

(vi)

\frac{- x + 2}{- 2} = \frac{y - 1}{7} = \frac{z + 3}{- 3} and \frac{x + 2}{- 1} = \frac{2 y - 8}{4} = \frac{z - 5}{4}

Q. Find the shortest distance between the following pairs of lines whose cartesian equations are:
(i)

\frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z - 3}{4} and \frac{x - 2}{3} = \frac{y - 3}{4} = \frac{z - 5}{5}

(ii)

\frac{x - 1}{2} = \frac{y + 1}{3} = z and \frac{x + 1}{3} = \frac{y - 2}{1}; z = 2

(iii)

\frac{x - 1}{- 1} = \frac{y + 2}{1} = \frac{z - 3}{- 2} and \frac{x - 1}{1} = \frac{y + 1}{2} = \frac{z + 1}{- 2}

(iv)

\frac{x - 3}{1} = \frac{y - 5}{- 2} = \frac{z - 7}{1} and \frac{x + 1}{7} = \frac{y + 1}{- 6} = \frac{z + 1}{1}

w T l_{2} O_{3} (s) + x N H_{2} O H (a q) \to y T l O H (s) + z N_{2} (g)

where

w, x, y, z

are the stoichiometric coefficients of

T l_{2} O_{3}, N H_{2} O H, T l O H, N_{2}

respectively.
Choose the correct statements:

Question 4 (a)

Identify terms which contain x and give the coefficient of x.

(i) $y^{2} x + y$

(ii) $13 y^{2} - 8 y x$

(iii) x + y + 2

(iv) 5 + z + zx

(v) 1 + x + xy

(vi) $12 x y^{2} + 25$

(vii) $7 x + x y^{2}$

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Cl2(g)+X−OH→YClO3−+ZH2O+10e−The coefficient X, Y and Z are:

The correct option is A 12, 2, 6Cl2+OH−→ClO−3+zH2OCl2→2ClO−3+12H++10e−+6H2OAdding 12OH− on both sides.Cl2+6H2O+12OH−→2ClO−3(12H−+12OH−)+10e−⇒Cl2+6H2O+12OH−→2ClO−3+12H2O+10e−CL2+12OH−→2ClO−3+6H2O+10e−x=12 y=2 z=6

${C l}_{2} (g) + X - O H \to Y C l {O_{3}}^{-} + Z H_{2} O + 10 e^{-}$
The coefficient X, Y and Z are: