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Structural Isomerism

Number of mol...

Question

Number of moles of $N H_{3}$ formed when $0.535 g$ of $N H_{4} C l$ is completely decomposed by $N a O H$ , is:
$N H_{4} C l + N a O H \to N H_{3} + N a C l + H_{2} O$

0.01 mol

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5.35 mol

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1.7 g

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0.17 mol

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Solution

The correct option is C 0.01 mol
According to the reaction,

53.5 g of $N H_{4} C l$ will completely react with 40 g $N a O H$

Hence, 0.535 g of $N H_{4} C l$ will completely react with 0.4 g $N a O H$

Hence number of moles of $N a O H = \frac{0.4}{40} = 0.01$

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Similar questions

N H_{3}

N H_{4} C l + N a C l

N a O H

N H_{4} C l + N a O H \to N H_{3} + N a C l + H_{2} O

A

N H_{4} C l

D

C O_{2}

B

C

+ N a O H \to N a C l + N H_{3} + H_{2} O

N H_{3} + C O_{2} + H_{2} O \to (B)

+ N a C l \to (C) + N H_{4} C l

\to N a_{2} C O_{3} + H_{2} O + (D)

N H_{4} C l

N a O H

H C_{2} H_{3} O_{2}

N a C l

N H_{3} + N H_{4} C l

N H_{3}

H C l

H C l O_{4}

K_{2} S O_{4}

N H_{4} C l

N H_{3}

C l_{2}

N H_{3} + C l_{2} \to N_{2} + N H_{4} C l

N H_{4} C l

N a O H

H C_{2} H_{3} O_{2}

N a C l

N H_{3} + N H_{4} C l

N H_{3}

H C l

H C l O_{4}

(N H_{4})_{2} S O_{4}

K_{2} S O_{4}

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