Dichromate ion in acidic medium oxidises stannous ion as :
$x S n^{2 +} + y C r_{2} O_{7}^{2 -} + z H^{+} \to a S n^{4 +} + b C r^{3 +} + c H_{2} O$

The value of

x : y

1 : 3

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The value of

x + y + z

18

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a : b

3 : 2

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The value of

z - c

7

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Solution

The correct option is D The value of $z - c$ is $7$
The balanced chemical reaction is :
$3 S n^{2 +} + y C r_{2} O_{7}^{2 -} + 14 H^{+} \to 3 S n^{4 +} + 2 C r^{3 +} + 7 H_{2} O$
(A) $x : y = \frac{3}{1} = 3 : 1$
(B) $x + y + z = 3 + 1 + 14 = 18$
(C) $\frac{a}{b} = \frac{3}{2} = 3 : 2$
(D) $z - c = 14 - 7 = 7$

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Similar questions

x S n^{2 +} + y C r_{2} O_{7}^{2 -} + z H^{+} \to a S n^{4 +} + b C r^{3 +} + c H_{2} O

$x S n^{2 +} + y C r_{2} O_{7}^{2 -} + z H^{+} \to a S n^{4 +} + b C r^{3 +} + c H_{2} O$

x S n^{2 +} + y C r_{2} O_{7}^{2 -} + z H^{+} \to a S n^{4 +} + b C r^{3 +} + c H_{2} O

Dichromate ion in acidic medium oxidizes stannous ion as

$x - - S n^{2 +} + y - C r_{2} O_{7}^{2 -} + z - H^{+} \to a - - S n^{4 +} + b - C r^{3 +} + c - H_{2} O$ then

x S n^{2 +} + y C r_{2} O_{7}^{2 -} + z H^{+} ⟶ a S n^{4 +} + b C r^{3 +} + c H_{2} O

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