bạn bấm máy tính thay m C2H5OH thêm = x 
16% là đề cho bạn

\(a.V_{\uparrow}=V_{CH_4}=2.22l\\ V_{C_2H_4}=6,72-2,22=4,5l\\ b.n_{CH_4}=\dfrac{2,22}{22,4}=\dfrac{111}{1120}mol\\ n_{C_2H_4}=\dfrac{4,5}{22,4}=\dfrac{45}{224}mol\\ CH_4+2O_2\xrightarrow[]{t^0}CO_2+2H_2O\\ C_2H_4+3O_2\xrightarrow[]{t^0}2CO_2+2H_2O\\ V_{O_2}=\left(\dfrac{111}{1120}\cdot2+\dfrac{45}{224}\cdot3\right)\cdot22,4=17,94l\)

\(n_{H_2}=\dfrac{9,916}{24,79}=0,4mol\\ n_{Zn}=a;n_{Al}=b\\ Zn+2HCl\xrightarrow[]{}ZnCl_2+H_2\\ 2Al+6HCl\xrightarrow[]{}2AlCl_3+3H_2\\ \Rightarrow\left\{{}\begin{matrix}a+1,5b=0,4\\65a+27b=11,9\end{matrix}\right.\\ \Rightarrow a=0,1;b=0,2\\ \%m_{Zn}=\dfrac{0,1.65}{11,9}\cdot100\%=54,62\%\\ \%m_{Al}=100\%-54,62\%=45,38\%\)

\(n_{C_2H_4}=\dfrac{2,8}{28}=0,1mol\\ \%V_{C_2H_4}=\dfrac{0,1.22,4}{11,2}\cdot100\%=20\%\\ \%V_{CH_4}=100\%-20\%=80\%\)

Khối lượng rượu có trong 400g dd \(C_2H_5OH\) 20% 

\(m_{C_2H_5OH}=\dfrac{400.20\%}{100\%}=80g\)

Khối lượng dd \(C_2H_5OH\) 8% cần thêm 

\(C_{\%C_2H_5OH}=\dfrac{80+\dfrac{m_{C_2H_5OH.thêm}.8\%}{100\%}}{m_{C_2H_5OH.thêm}+400}\cdot100\%=16\%\\ \Rightarrow m_{C_2H_5OH.thêm}=200g\)

\(a.n_{H_2}=\dfrac{3,7185}{24,79}=0,15mol\\ Mg+2HCl\rightarrow MgCl_2+H_2\)

0,15        0,3                0,15              0,15

\(\%m_{Mg}=\dfrac{0,15.24}{96}\cdot100\%=3,75\%\\ \%m_{MgO}=96,25\%\\ b.n_{MgO}=\dfrac{96.96,25\%}{40}=2,31mol\\ MgO+2HCl\rightarrow MgCl_2+H_2O\)

2,31           4,62              2,31                  2,31

\(C_{\%HCl}=\dfrac{\left(0,3+4,62\right).36,5}{150}\cdot100\%=119,72\%?\\ c.C_{\%MgCl_2}=\dfrac{\left(0,15+2,31\right).95}{96+150-0,15.2}\cdot100\%=95,12\%\)

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B cùng có hóa trị II