In the chemical reaction,
$K_{2} {C r}_{2} O_{7} + x H_{2} S O_{4} + y S O_{2} ⟶ K_{2} S O_{4} + {C r}_{2} {(S O_{4})}_{3} + z H_{2} O$
$x, y$ and $z$ are, respectively :

1, 3, 1

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4, 1, 4

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3, 2, 3

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2, 1, 2

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Solution

The correct option is A $1, 3, 1$
The balanced equation is given below:
$14 H^{\oplus} + 6 e^{-} + {C r}_{2} O_{7}^{2 -} ⟶ 2 {C r}^{3 +} + 7 H_{2} O$
$6 H_{2} O + 3 S O_{2} ⟶ 3 S O_{4}^{2 -} + 6 e^{-} + 12 H^{\oplus}$
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${C r}_{2} O_{7}^{2 -} + 2 H^{\oplus} + 3 S O_{2} ⟶ 2 {C r}^{3 +} + 3 S O_{4}^{2 -} + H_{2} O$
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It can be written as follows:
$K_{2} {C r}_{2} O_{7} + H_{2} S O_{4} + 3 S O_{2} ⟶ {C r}_{2} {(S O_{4})}_{3} + K_{2} S O_{4} + H_{2} O$
$∴ x = 1, y = 3, z = 1$

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