Viết PTHH hoàn thành chuỗi biến hoá sau : Đồng \(\buildrel {(1)} \over\longrightarrow \) đồng (II) oxit\(\buildrel {(2)} \over\longrightarrow \) đồng (II) clorua\(\buildrel {(3)} \over\longrightarrow \) đồng (I) clorua \(\buildrel {(4)} \over\longrightarrow \) đồng (II) clorua \(\buildrel {(5)} \over \longrightarrow \) đồng (II) hidroxit \(\buildrel {(6)} \over\longrightarrow \) đồng (II) nitrat \(\buildrel {(7)} \over\longrightarrow \) khí nito (IV) oxit Hướng dẫn trả lời: \(\left( 1 \right){\rm{ }}Cu{\rm{ }} + {\rm{ }}{O_2}\buildrel {{t^0}} \over \longrightarrow 2CuO\) (2) CuO + 2HCl → CuCl2 + H2O (3) CuCl2 + Cu → 2CuCl \(\left( 4 \right){\rm{ }}2CuCl{\rm{ }} + {\rm{ }}C{l_2}\buildrel {{t^0}} \over \longrightarrow 2CuC{l_2}\) (5) CuCl2 + NaOH → Cu(OH)2 ↓+ 2NaCl (6) Cu(OH)2 + 2HNO3→Cu(NO3)2 + 2H2O \(\left( 7 \right)Cu{\left( {N{O_3}} \right)_2}\buildrel {{t^0}} \over \longrightarrow CuO{\rm{ }} + {\rm{ }}2N{O_2} \uparrow + {1 \over 2}{O_2} \uparrow \).
