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Redox Reactions

The following...

Question

The following series of reactions show the formation of $N a C l O_{3}$ .

$i) K M n O_{4} + H C l \to K C l + M n C l_{2} + H_{2} O + C l_{2}$
$i i) C l_{2} + B a (O H)_{2} \to B a (C l O_{3})_{2} + B a C l_{2} + H_{2} O$
$i i i) B a (C l O_{3})_{2} + N a_{2} S O_{4} \to B a S O_{4} + N a C l O_{3}$
Pick the incorrect statement(s) about this process:

1 eq

H C l

produces 1 eq of

N a C l O_{3}

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1 mol

H C l

produces 5/48 mol of

N a C l O_{3}

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n-factor of

K M n O_{4}

in reaction (i) is 3

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1 mol of

H C l

produces 5/16 mol of

C l_{2}

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Solution

The correct options are
A 1 eq $H C l$ produces 1 eq of $N a C l O_{3}$
C n-factor of $K M n O_{4}$ in reaction (i) is 3
Balanced reactions are:

$i) 2 K M n O_{4} + 16 H C l \to 2 K C l + 2 M n C l_{2} + 8 H_{2} O + 5 C l_{2}$

$i i) 6 C l_{2} + 6 B a (O H)_{2} \to B a (C l O_{3})_{2} + 5 B a C l_{2} + 6 H_{2} O$

$i i i) B a (C l O_{3})_{2} + N a_{2} S O_{4} \to B a S O_{4} + 2 N a C l O_{3}$

Here, $1 H C l \to \frac{5}{16} C l_{2} \frac{5}{16} C l_{2} \to \frac{2}{6} \times \frac{5}{16} N a C l O_{3}$

$i) M n^{7 +} \to M n^{2 +}$
n-factor of $K M n O_{4}$ is 5.

Suggest Corrections

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