You visited us 1 times! Enjoying our articles? Unlock Full Access!

Stoichiometric Calculations

Consider the ...

Question

Consider the following reactions:
Path (I)
$N a N H_{2} + N_{2} O \to A + N H_{3} + N a O H$
$A + H_{2} S O_{4} \to B + N a_{2} S O_{4}$
Path (II)
$N a N O_{2} + H_{2} S O_{4} i n s i t u - --- \to C + N a_{2} S O_{4}$
$C + N_{2} H_{4} \to B + H_{2} O$
The two paths are used to form compound B, which is used in the Schmidt reaction. If we start with 0.2 moles of $N_{2} O$ in Path (I) and 0.6 moles of $N a N O_{2}$ in Path (II) to form compound B, then $x$ millimoles of B are formed. The value of $x$ is

Open in App

Solution

$2 N a N H_{2} + N_{2} O \to N a N_{3} + N H_{3} + N a O H$ and $N a N_{3} + \frac{1}{2} H_{2} S O_{4} \to H N_{3} + \frac{1}{2} N a_{2} S O_{4}$

$N a N O_{2} + \frac{1}{2} H_{2} S O_{4} i n s i t u - ---- \to H N O_{2} + \frac{1}{2} N a_{2} S O_{4}$ and $H N O_{2} + N_{2} H_{4} \to H N_{3} + 2 H_{2} O$

From the balanced reaction, 0.2 moles of $N_{2} O$ will produce 0.2 moles of $H N_{3}$ , and 0.6 moles of $N a N O_{2}$ will produce 0.6 moles of $H N_{3}$ (B).
Hence, total moles of B formed = 0.8 moles = 800 mmoles.

Suggest Corrections

Similar questions

N a N H_{2} + N_{2} O \to A + N H_{3} + N a O H

A + H_{2} S O_{4} \to B + N a_{2} S O_{4}

N a N O_{2} + H_{2} S O_{4} i n s i t u - --- \to C + N a_{2} S O_{4}

C + N_{2} H_{4} \to B + H_{2} O

N_{2} O

N a N O_{2}

x

x

H_{2} S O_{4}

C O_{2}

H_{2} O .

H_{2} {SO}_{4}

Ni

{CO}_{2}

H_{2} O

A \to B, B \to 2 C, 3 C \to 4 D

D

A

Join BYJU'S Learning Program