${\text{n}}_{{\text{NH}}_3}=0,6 \text{mol};{\text{n}}_{{\text{CO}}_2}=0,5 \text{mol}$

Gọi x, y lần lượt là số mol của NH₄HCO₃ và (NH₄)₂CO₃ trong hỗn hợp ban đầu

$$\begin{gathered} \begin{array}{l}{\text{NH}}_4{\text{HCO}}_3\stackrel{t^o}{\to }{\text{NH}}_3+{\text{CO}}_2+{\text{H}}_2\text{O}\\ \text{x\hspace{0.17em}mol \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}x\hspace{0.17em}mol \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}x\hspace{0.17em}mol\hspace{0.17em}}\end{array} \\ \begin{array}{l}{\left({\text{NH}}_4\right)}_2{\text{CO}}_3\stackrel{{\text{t}}^0}{\to }2{\text{NH}}_3+{\text{CO}}_2+{\text{H}}_2\\ \text{y\hspace{0.17em}mol \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}2\text{y\hspace{0.17em}mol\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}y\hspace{0.17em}mol}\end{array} \end{gathered}$$

Ta có: $\left\{\begin{array}{l}x+2y=0,6\\ x+y=0,5\end{array}\to \left\{\begin{array}{l}x=0,4\\ y=0,1\end{array}\right.\right.$