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Double Displacement Reaction

Double displa...

Question

Double displacement reactions are :

M g + C u S O_{4} \to M g S O_{4} + C u

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K O H + H N O_{3} \to H_{2} O + K N O_{3}

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N a_{2} S + 2 H C I \to 2 N a C l + H_{2} S

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

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Solution

The correct options are
B $K O H + H N O_{3} \to H_{2} O + K N O_{3}$
C $N a_{2} S + 2 H C I \to 2 N a C l + H_{2} S$
D $B a C l_{2} + H_{2} S O_{4} \to B a S O_{4} + 2 H C l$
Those reactions in which two compounds react by an exchange of ions to form two new compounds are called double displacement reactions. In double replacement reactions, the positive ions and negative ions switch their positions and form two new compounds or products.
Hence;
$K O H + H N O_{3} \to H_{2} O + K N O_{3}$
$N a_{2} S + 2 H C I \to 2 N a C l + H_{2} S$
$B a C l_{2} + H_{2} S O_{4} \to B a S O_{4} + 2 H C l$
are double displacement reactions.

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Similar questions

A g N O_{3} + N a B r \to N a N O_{3} + A g B r

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

A g N O_{3} + N a C l \to A g C l + N a N O_{3}

N a O H + H C l \to N a C l + H_{2} O

F e + 2 H C l \to F e C l_{2} + H_{2}

F e S + 2 H C l \to F e C l_{2} + H_{2} S

F e + S Δ - \to F e S

N a C l + H_{2} S O_{4} < 200^{o} C - --- \to N a H S O_{4} + H C l

C u S O_{4} + H_{2} S \to C u S + H_{2} S O_{4}

S + H_{2} S O_{4} \to 3 S O_{2} + H_{2} O

P b + C u C l_{2} \to P b C l_{2} + C u

N a_{2} S O_{4} + B a C l_{2} \to B a S O_{4} + 2 N a C l

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

M n O_{2} + H C l \to M n C l_{2} + C l_{2} + H_{2} O

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B a C l_{2} + A l_{2} (S O_{4})_{3} \to B a S O_{4} + A l C l_{3}

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Z n + H N O_{3} \to Z n (N O_{3})_{2} + N O_{2} + H_{2} O

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P b O + C \to P b + C O_{2}

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C H_{4} + O_{2} \to C O_{2} + H_{2} O

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