Lập PTHH có sơ đồ phản ứng sau
1, M+H2O –> M(OH)n +H2
2,M+HCl —>………..+H2
3, M+H2SO4 —> M2(SO4)n +SO2+H2O
4,CxHy+O2 —> CO2+H2O
5,CxHyOz+O2—> CO2+H2O
6,FexOy_HCl—>………+H2O
7,M+HCl—>………+H2
8, M+H2SO4—-> …….+H2
9,M+O2—>……..
10, M+HCl—>………….
11,MxOy+H2—->………….
12,M+H2O—->……………
Lập PTHH có sơ đồ phản ứng sau 1, M+H2O –> M(OH)n +H2 2,M+HCl —>………..+H2 3, M+H2SO4 —> M2(SO4)n +SO2+H2O 4,CxHy+O2 —> CO2+H2O 5,CxHyOz+O2
By Amara
Bài giải:
1. $2M+nH_2O→2M(OH)_n+nH_2↑$
2. $2M+2nHCl→2MCl_n+nH_2↑$
3. $2M+2nH_2SO_4→M_2(SO_4)_n+nSO_2↑+2nH_2O$
4. $4C_xH_y+(4x+y)O_2\xrightarrow{t^o} 4xCO_2↑+2yH_2O$
5. $4C_xH_yO_z+(4x+y-2z)O_2\xrightarrow{t^o} 4xCO_2↑+2yH_2O$
6. $Fe_xO_y+2yHCl→xFeCl_{\frac{2y}{x}}+yH_2O$
7. $2M+2nHCl→2MCl_n+nH_2↑$
8. $2M+nH_2SO_4→M_2(SO_4)_n+nH_2↑$
9. $4M+nO_2\xrightarrow{t^o}2M_2O_n $
10. $2M+2nHCl→2MCl_n+nH_2↑$
11. $M_xO_y+yH_2\xrightarrow{t^o}xM↓+yH_2O $
12. $2M+nH_2O→2M(OH)_n+nH_2↑$
Em tham khảo nha:
\(\begin{array}{l}
1)\\
2M + 2n{H_2}O \to 2M{(OH)_n} + n{H_2}\\
2)\\
2M + 2nHCl \to 2MC{l_n} + n{H_2}\\
3)\\
M + 2n{H_2}S{O_4} \to {M_2}{(S{O_4})_n} + nS{O_2} + 2n{H_2}O\\
4)\\
2{C_x}{H_y} + \frac{{4x + y}}{2}{O_2} \to 2xC{O_2} + y{H_2}O\\
5)\\
2{C_x}{H_y}{O_z} + \frac{{4x + y – 2z}}{2}{O_2} \to 2xC{O_2} + y{H_2}O\\
6)\\
F{e_x}{O_y} + 2yHCl \to xFeC{l_{\frac{{2y}}{x}}} + y{H_2}O\\
7)\\
2M + 2nHCl \to 2MC{l_n} + n{H_2}\\
8)\\
2M + n{H_2}S{O_4} \to {M_2}{(S{O_4})_n} + n{H_2}\\
9)\\
4M + n{O_2} \to 2{M_2}{O_n}\\
10)\\
2M + 2nHCl \to 2MC{l_n} + n{H_2}\\
11)\\
{M_x}{O_y} + y{H_2} \to xM + y{H_2}O\\
12)\\
2M + 2n{H_2}O \to 2M{(OH)_n} + n{H_2}
\end{array}\)