Hướng dẫn giải

PTHH: $CuC{{l}_{2}}+2KOH\to Cu{{\left( OH \right)}_{2}}\downarrow +2NaCl$

         0,02$\left( mol \right)$   1 $\left( mol \right)$    0,02 $\left( mol \right)$   0,4 $\left( mol \right)$.

$FeC{{l}_{3}}+3KOH\to Fe{{\left( OH \right)}_{3}}\downarrow +3KCl$

0,01$\left( mol \right)$$\to $0,03$\left( mol \right)$   0,01$\left( mol \right)$

$Cu{{\left( OH \right)}_{2}}\xrightarrow{{{t}^{o}}}CuO+{{H}_{2}}O$         $2Fe{{\left( OH \right)}_{3}}\to F{{e}_{2}}{{O}_{3}}+3{{H}_{2}}O$

0,02$\left( mol \right)$ $\to $0,02 $\left( mol \right)$                  0,01$\left( mol \right)$$\to $0,005$\left( mol \right)$

${{m}_{CuO}}={{n}_{CuO}}\times M=0,02\times 80=1,6\left( gam \right).$

${{m}_{F{{e}_{2}}{{O}_{3}}}}={{n}_{F{{e}_{2}}{{O}_{3}}}}\times M=0,005\times 160=0,8\left( gam \right).$

Khối lượng chất rắn sau khi nung: ${{m}_{hh}}={{m}_{CuO}}+{{m}_{F{{e}_{2}}{{O}_{3}}}}=1,6+0,8=2,4\left( gam \right)$.

Chọn đáp án C.