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Oxides of Sulphur

BaS+2HCl→ ?+B...

Question

$B a S + 2 H C l \to ? + B a C l_{2}$

H_{2} S

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H_{2} S O_{4}

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H_{2} C O_{3}

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S O_{2}

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Solution

The correct option is A $H_{2} S$
$B a s + 2 H C l \to H_{2} S + B a C l_{2}$
When barium sulfide reacts with hydrogen chloride it forms hydrogen sulfide ( $H_{2} S$ ) and barium chloride
Hence option A is correct.

Suggest Corrections

Similar questions

B a C l_{2} + H_{2} S O_{4} \to B a S O_{4} + 2 H C l

B a C l_{2}

H_{2} S O_{4}

B a S O_{4}

(B a = 137, C l = 35.5, S = 32, H = 1, O = 16)

B a C l_{2} + H_{2} S O_{4} \to B a S O_{4} + 2 H C l

B a C l_{2} + H_{2} S O_{4} \to B a S O_{4} + 2 H C l

S + O_{2} \to S O_{2} (Yield = 80%)

S O_{2} + C l_{2} + 2 H_{2} O \to H_{2} S O_{4} + 2 H C l (Yield = 60%)

H_{2} S O_{4} + B a C l_{2} \to B a S O_{4} + 2 H C l (Yield = 30%)

S O_{2}

C l_{2}

B a C l_{2}

B a S O_{4}

137 g/mol

H_{2} S

1.69

B a S

B a C l_{2}

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