$M g S O_{4}$ on reaction with $N H_{4} O H$ and $N a_{2} H P O_{4}$ forms a white crystalline precipitate. What is its formula?

M g (N H_{4}) P O_{4}

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M g_{3} (P O_{4})_{2}

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M g (O H)_{2}

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M g S O_{4}

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Solution

The correct option is A $M g (N H_{4}) P O_{4}$
$M g S O_{4}$ on reaction with $N H_{4} O H$ and $N a_{2} H P O_{4}$ forms a white crystalline precipitate. The formula of precipitate is $M g (N H_{4}) P O_{4}$
$M g S O_{4} + N a_{2} H P O_{4} + N H_{4} O H \to M g (N H_{4}) P O_{4} + N a_{2} S O_{4} + H_{2} O$ .
Hence option A is correct.

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Similar questions

M g S O_{4}

N H_{4} O H

N a_{2} H P O_{4}

$M g S O_{4}$ on reaction with $N H_{4} O H$ and $N a_{2} H P O_{4}$ forms a white crystalline precipitate. What is its formula?

M g S O_{4} + N H_{4} O H + N a_{2} H P O_{4} \to

M g S O_{4}

N H_{4} C I

N a_{2} H P O_{4}

M g S O_{4}

N H_{4} C l

N a_{3} P O_{4}

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