Đặt : 

nC2H5OH = a (mol) 

nCH3OH = b (mol) 

=> mhh = 46a + 32b = 5.5 (g) (1) 

nH2 = 0.075 (mol) 

C2H5OH + K => C2H5OK + 1/2 H2 

CH3OH + K => CH3OK + 1/2H2

=> 0.5a + 0.5b = 0.075 (2) 

(1) , (2) : 

a = 0.05

b = 0.1 

%m C2H5OH = 0.05*46/5.5 * 100% = 41.82%

%m CH3OH = 58.18%

\(2Na+2H_2O\rightarrow2NaOH+H_2\\ Na_2O+H_2O\rightarrow2NaOH\\ n_{H_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\\ n_{Na}=2.0,3=0,6\left(mol\right)\\ a,m_{Na}=0,6.23=13,8\left(g\right)\\ m_{Na_2O}=26,2-13,8=12,4\left(g\right)\\b, n_{Na_2O}=\dfrac{12,4}{62}=0,2\left(mol\right)\\ n_{NaOH\left(tổng\right)}=n_{Na}+2.n_{Na_2O}=0,6+\dfrac{12,4}{62}=0,8\left(mol\right)\\ m_{c.tan}=m_{NaOH}=0,8.40=32\left(g\right)\\ c,m_{ddNaOH}=m_{hh}+m_{H_2O}-m_{H_2}=26,2+200-0,3.2=225,6\left(g\right)\\ C\%_{ddNaOH}=\dfrac{32}{225,6}.100\approx14,185\%\)

Anh có làm rồi nha!

\(n_{H_2}=\dfrac{13,44}{22,4}=0,6\left(mol\right)\)

\(2A1+2NAOH+2H_2O-2NaA10_2+H_2O\)

\(AI_2O_3=2NaOH+2NaOHA10_2+H_2O\)

\(n_{AI}=\dfrac{2}{3}n_{H_2}=\dfrac{2}{3}.0,6=0,4\left(mol\right)\)

\(m_{AI}=27.0,4=10,8\left(gam\right);mAI_2O_3=31,2-10,8=20,4\left(gam\right)\)

Biết làm mỗi câu A

 Chỉ Có Al tác dụng với HCl sp tạo ra khí H₂

\(n_{H_2}=\dfrac{13.44}{22.4}=0.6\left(mol\right)\)
PTHH:
\(2Al+6HCl\rightarrow2AlCl_3+3H_2\)
   0.4           \(\leftarrow\)                           0.6  (mol)

\(Al_2O_3+HCl\rightarrow AlCl_3+H_2O\)

n_Al=\(\dfrac{0.6\cdot2}{3}=0.4\left(mol\right)\)

m_Al=0.4*27=10.8(g)

=)) \(m_{Al_2O_3}=31.2-10.8=20.4\left(g\right)\)

\(\left\{{}\begin{matrix}\%Al=\dfrac{10.8}{31.2}\cdot100=34.62\%\\\%Al_2O_3=\dfrac{20.4}{31.2}\cdot100=65.38\%\end{matrix}\right.\)

\(n_{H_2}=\dfrac{7,84}{22,4}=0,35mol\)

Gọi \(\left\{{}\begin{matrix}n_{Fe}=x\\n_{Zn}=y\end{matrix}\right.\)

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

 x                                     x ( mol )

\(Zn+2HCl\rightarrow ZnCl_2+H_2\)

 y                                     y      ( mol )

Ta có:

\(\left\{{}\begin{matrix}56x+65y=21,4\\x+y=0,35\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=0,15\\y=0,2\end{matrix}\right.\)

\(\Rightarrow m_{Fe}=0,15.56=8,4g\)

\(\Rightarrow m_{Zn}=0,2.65=13g\)

\(\%m_{Fe}=\dfrac{8,4}{21,4}.100=39,25\%\)

\(\%m_{Zn}=100\%-39,25\%=60,75\%\)

\(m_{FeCl_2}=0,15.127=19,05g\)

\(m_{ZnCl_2}=0,2.136=27,2g\)

\(Đặt:n_{MnO_2}=a\left(mol\right),n_{KMnO_4}=b\left(mol\right)\)

\(m_{hh}=87a+158b=37.96\left(g\right)\left(1\right)\)

\(n_{Cl_2}=\dfrac{10.08}{22.4}=0.45\left(mol\right)\)

\(2KMnO_4+16HCl\rightarrow2KCl+2MnCl_2+5Cl_2+8H_2O\)

\(MnO_2+4HCl\rightarrow MnCl_2+Cl_2+2H_2O\)

\(n_{Cl_2}=a+2.5b=0.45\left(mol\right)\left(2\right)\)

\(\left(1\right),\left(2\right):a=0.4,b=0.02\)

\(\%MnO_2=\dfrac{0.4\cdot87}{37.96}\cdot100\%=91.68\%\\\%KMnO_4=100-91.68=8.32\% \)

\(m_M=m_{KCl}+m_{MnCl_2}=0.02\cdot74.5+\left(0.4+0.02\right)\cdot126=54.41g\)

Theo đề bài ta có hệ phương trình:

Theo đề bài ta có hệ phương trình: