Nung nóng m (gam) $\text{Cu}{\left({\text{NO}}_3\right)}_2$ sau một thời gian dừng lại, làm nguội rồi cân thấy khối lượng giảm 0,54 gam. Khối lượng $\text{Cu}{\left({\text{NO}}_3\right)}_2$ đã bị nhiệt phân là

A. 0,49 gam.                                                 

Ta có: $\begin{array}{l}\text{Cu}{\left({\text{NO}}_3\right)}_2\stackrel{t^0}{\to }\text{CuO}+2{\text{NO}}_2+0,5{\text{O}}_2\\ \text{a\hspace{1em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}2a0,5a\end{array}$

$\to m_{\text{giam }}=m_{{\text{NO}}_2}+m_{{\text{O}}_3}=2a.46+0,5a.32=0,54\left(\text{gam}\right)$

$\to a=0,005 \text{mol\hspace{1em}}\to m_{\text{Cu}{\left({\text{NO}}_3\right)}_2}=0,94\left(\text{gam}\right)$