`@` Fe:

\(4FeS_2+11O_2\rightarrow\left(t^o\right)2Fe_2O_3+8SO_2\) 

\(2NaCl+2H_2O\xrightarrow[cmn]{đpdd}2NaOH+Cl_2+H_2\) 

\(Fe_2O_3+3H_2\rightarrow\left(t^o\right)2Fe+3H_2O\) 

`@` \(Fe\left(OH\right)_3\):

\(2SO_2+O_2⇌\left(t^o,V_2O_5\right)2SO_3\)

\(SO_3+H_2O\rightarrow H_2SO_4\)

\(Fe+H_2SO_4\rightarrow FeSO_4+H_2\)

\(FeSO_4+2NaOH\rightarrow Fe\left(OH\right)_2\downarrow+Na_2SO_4\)

`@` \(FeCl_2\) :

\(H_2+Cl_2\xrightarrow[as]{t^o}2HCl\)

\(Fe+2HCl\rightarrow FeCl_2+H_2\)

`@` \(FeCl_3\) :

\(Fe_2O_3+6HCl\rightarrow2FeCl_3+3H_2O\)

:)) e đọc đề hongg kĩ:v

\(n_X=n_{O_2}=\dfrac{0,224}{32}=0,007\left(mol\right)\)

\(M_X=\dfrac{0,42}{0,007}=60\) \((g/mol)\)

\(PTHH:X+O_2\rightarrow\left(t^o\right)CO_2+H_2O\)

\(m_{O_2}=32.\dfrac{10,08}{22,4}=14,4\left(g\right)\)

\(BTKL:m_X+m_{O_2}=m_{CO_2}+m_{H_2O}\)

\(\Rightarrow m_{CO_2}+m_{H_2O}=6+14,4=20,4\left(g\right)\) (1)

Ta có: \(\dfrac{m_{CO_2}}{m_{H_2O}}=\dfrac{11}{6}\)  \(\Leftrightarrow6m_{CO_2}-11m_{H_2O}=0\) (2)

\(\left(1\right);\left(2\right)\Rightarrow\left\{{}\begin{matrix}m_{CO_2}=13,2\\m_{H_2O}=7,2\end{matrix}\right.\)

Bảo toàn C: \(n_C=n_{CO_2}=\dfrac{13,2}{44}=0,3\left(mol\right)\)

Bảo toàn H: \(n_H=2.n_{H_2O}=2.\dfrac{7,2}{18}=0,8\left(mol\right)\)

\(n_O=\dfrac{6-\left(0,3.12+0,8.1\right)}{16}=0,1\left(mol\right)\)

Đặt CTTQ X: \(C_xH_yO_z\)

\(x:y:z=0,3:0,8:0,1=3:8:1\)

\(CTĐG:\left(C_3H_8O\right)_n=60\)

                 \(\Rightarrow n=1\)

\(\Rightarrow CTPTX:C_3H_8O\)

So sánh sự giống nhau và khác nhau của nguyên phân và giảm phân

a.\(\left(x-\dfrac{x^2+x^2}{x+y}\right).\left(\dfrac{1}{y}+\dfrac{2}{x-y}\right)\)

\(=\left(\dfrac{x^2+xy-2x^2}{x+y}\right)\left(\dfrac{x-y+2y}{y\left(x-y\right)}\right)\)

\(=\left(\dfrac{xy-x^2}{x+y}\right)\left(\dfrac{x+y}{y\left(x-y\right)}\right)\)

\(=-\left(\dfrac{x\left(x-y\right)}{x+y}\right)\left(\dfrac{x+y}{y\left(x-y\right)}\right)\)

\(=-\dfrac{x}{y}\)

b.\(\dfrac{2}{xy}:\left(\dfrac{1}{x}-\dfrac{1}{y}\right)-\dfrac{x+y}{\left(x-y\right)^2}\)

\(=\dfrac{2}{xy}:\left(\dfrac{y-x}{xy}\right)-\dfrac{x+y}{\left(x-y\right)^2}\)

\(=\dfrac{2}{xy}.\dfrac{xy}{y-x}-\dfrac{x+y}{\left(x-y\right)^2}\)

\(=-\dfrac{2}{x-y}-\dfrac{x+y}{\left(x-y\right)^2}\)

\(=\dfrac{2x+2y-x-y}{\left(x-y\right)^2}\)

\(=\dfrac{x+y}{\left(x-y\right)^2}\)