Lập luận tìm x y a) FeClx + Cl2 -> FeCly b) CxHyO + O2 -> 2CO2 + 3H2O c) KMnO4 + HCl -> MnCl2 + KxCly + H2O + Cl2

Em tham khảo nha :

\(\begin{array}{l}
a)\\
FeC{l_x} + C{l_2} \to FeC{l_y}\\
C{l^0} + 1e \to C{l^ – }\\
F{e^{ + x}} \to F{e^{ + y}} + (y – x)e\\
 \Rightarrow y > x\\
 \Rightarrow x = 2;y = 3\\
2FeC{l_2} + C{l_2} \to 2FeC{l_3}\\
b)\\
{C_x}{H_y}O + {O_2} \to 2C{O_2} + 3{H_2}O\\
{n_C} = {n_{C{O_2}}}\\
{n_H} = 2{n_{{H_2}O}}\\
 \Rightarrow x = \dfrac{{{n_C}}}{{{n_{{C_x}{H_y}O}}}} = \dfrac{{{n_{C{O_2}}}}}{{{n_{{C_x}{H_y}O}}}} = \dfrac{{2{n_{{C_x}{H_y}O}}}}{{{n_{{C_x}{H_y}O}}}} = 2\\
y = \dfrac{{{n_H}}}{{{n_{{C_x}{H_y}O}}}} = \dfrac{{2{n_{{H_2}O}}}}{{{n_{{C_x}{H_y}O}}}} = \dfrac{{6{n_{{C_x}{H_y}O}}}}{{{n_{{C_x}{H_y}O}}}} = 6\\
{C_2}{H_6}O + 3{O_2} \to 2C{O_2} + 3{H_2}O\\
c)\\
KMn{O_4} + HCl \to MnC{l_2} + {K_x}C{l_y} + C{l_2} + {H_2}O\\
K(I),Cl(I)\\
 \Rightarrow x = 1;y = 1\\
 \Rightarrow CTHH:KCl
\end{array}\)