Ecuația reacției chimice:
0.4 x
2NH₄Cl + Ca(OH)₂ -----> CaCl₂ + 2NH₃ + 2H₂O
2 2
Rezolvarea problemei:
[tex]m_{_{NH_4Cl}}=21.4\ g\\M_{_{NH_4Cl}}=A_N+4*A_H+A_{Cl}=14+4+35.5=53.5\ g/mol\\n=\frac{m}{M}=\frac{21.4}{53.5}=0.4\ moli\ NH_4Cl\\\\n_{_{NH_3}}=x=0.4\ moli\\{\eta}=\frac{Cp}{Ct}*100\implies Ct=\frac{Cp*100}{\eta}=\frac{0.4*100}{80}=0.5\ moli\ NH_3\\\\V_M=22.4\ dm^3\\n=\frac{V}{V_M}\implies V=n*V_M=0.5*22.4=11.2\ dm^3\ NH_3[/tex]
