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Atomic Number and Mass Number

Number of mol...

Question

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

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Solution

$N H_{4} C l + N a O H ⟶ N H_{3} + N a C l + H_{2} O$
$0.0535 g$ of $N H_{4} C l$ contains, $\frac{0.535}{53.5} m o l e$ of $N H_{4} C l$ .
As $1 m o l e$ of $N H_{4} C l$ gives $1 m o l e$ of $N H_{3}$ .
Thus, $\frac{0.535}{53.5} = 0.01 m o l e s$ of $N H_{4} C l$ will give $0.01 m o l e$ of $N H_{3}$ .
Thus, $0.01 m o l e s$ of $N H_{3}$ produced by $0.535 g$ of $N H_{4} C l$ .

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Similar questions

N H_{3}

0.535 g

N H_{4} 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 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}

N a C l + H_{2} O + C O_{2} + N H_{3}

\to

N H_{4} C l + X

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