Đáp án:

Giải thích các bước giải:

$1/M_{FeCl_2}=56+2.35,5=127(g/mol)\\M_{CuSO_4}=64+32+4.16=160(g/mol)\\\rightarrow C.\\2/M_{CuO}=80(g/mol)\\m=n.M=0,2.80=16(g)\\\rightarrow B.\\3/V=n.22,4=0,6.22,4=13,44(l)\\\rightarrow C.\\4/M_{CuSO_4}=64+32+4.16=160(g/mol)\\n=\dfrac{m}{M}=\dfrac{16}{160}=0,1(mol)\\\rightarrow A.\\6/d_{A/B}=44/2=22\\\rightarrow A.\\d_{NO_2/kk}=\dfrac{46}{29}~~1,6>1\\\rightarrow Loại\\d_{CO_2/kk}=\dfrac{44}{29}=1,57>1\\\rightarrow Loại\\d_{SO_2/kk}=\dfrac{64}{29}=2,02689>1\\\rightarrow Loại\\d_{H_2/kk}=\dfrac{1}{29}=0,0344827<1\\\rightarrow Chọn\\\rightarrow D.\\8/\%N=\dfrac{m_N}{M_{N_2O}}.100\%=\dfrac{2.14}{28+16}.100=63,6\%\\\rightarrow A.$

`#An`