Consider the following reactions: Path (I) NaNH2+N2O→A+NH3+NaOH A+H2SO4→B+Na2SO4 Path (II) NaNO2+H2SO4in situ−−−−→C+Na2SO4 C+N2H4→B+H2O The two paths are used to form compound B, which is used in the Schmidt reaction. If we start with 0.2 moles of N2O in Path (I) and 0.6 moles of NaNO2 in Path (II) to form compound B, then x millimoles of B are formed. The value of x is

Consider the following reactions: Path (I) NaNH2+N2O→A+NH3+NaOH A+H2

1.	Consider the following reactions: Path (I) NaNH2+N2O→A+NH3+NaOH A+H2SO4→B+Na2SO4 Path (II) NaNO2+H2SO4in situ−−−−→C+Na2SO4 C+N2H4→B+H2O The two paths are used to form compound B, which is used in the Schmidt reaction. If we start with 0.2 moles of N2O in Path (I) and 0.6 moles of NaNO2 in Path (II) to form compound B, then x millimoles of B are formed. The value of x is
Answer» 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