\(x+\dfrac{x}{5}+x\times\dfrac{1}{5}=\dfrac{1}{5}\)

\(\Rightarrow x\left(1+\dfrac{1}{5}+\dfrac{1}{5}\right)=\dfrac{1}{5}\)

\(\Rightarrow\dfrac{7}{5}x=\dfrac{1}{5}\)

`=>x=1/7`

Số mol Ba(OH)₂ ban đầu là 0,4 mol mà pt (1) p.ứ 0,1 mol nên còn 0,3 mol á bạn

\(n_{H_2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\)

\(Fe+2HCl\rightarrow FeCl_2+H_2\) (1)

0,1      0,2            0,1       0,1     ( mol )

\(Fe_2O_3+6HCl\rightarrow2FeCl_3+3H_2O\) (2)

\(m_{Fe}=0,1.56=5,6\left(g\right)\)

\(\left\{{}\begin{matrix}\%m_{Fe}=\dfrac{5,6}{8}.100=70\%\\\%m_{Fe_2O_3}=100\%-70\%=30\%\end{matrix}\right.\)

\(n_{Fe_2O_3}=\dfrac{8-5,6}{160}=0,015\left(mol\right)\)

\(\left(2\right)\Rightarrow n_{HCl}=6n_{Fe_2O_3}=6.0,015=0,09\left(mol\right)\)

\(\Sigma n_{HCl}=0,09+0,2=0,29\left(mol\right)\)

\(m_{dd\left(HCl\right)}=\dfrac{0,29.36,5}{10\%}=105,85\left(g\right)\)

\(V_{dd\left(HCl\right)}=\dfrac{105,85}{1,1}=96,22\left(ml\right)=0,09622\left(l\right)\)

\(\left\{{}\begin{matrix}C_{M_{FeCl_2}}=\dfrac{0,1}{0,09622}=1,03\left(M\right)\\C_{M_{FeCl_3}}=\dfrac{0,03}{0,09622}=0,31\left(M\right)\end{matrix}\right.\)

\(n_{Ba\left(OH\right)_2}=0,4.1=0,4\left(mol\right)\) ; \(n_{BaCO_3}=\dfrac{19,7}{197}=0,1\left(mol\right)\)

Nung A lại thu được kết tủa `->` dd A có \(Ba\left(HCO_3\right)_2\)

\(Ba\left(OH\right)_2+CO_2\rightarrow BaCO_3\downarrow+H_2O\)

   0,1              0,1           0,1                   ( mol )

\(Ba\left(OH\right)_2+2CO_2\rightarrow Ba\left(HCO_3\right)_2\)

    0,3             0,6                0,3        ( mol )

\(Ba\left(HCO_3\right)_2\rightarrow\left(t^o\right)BaCO_3\downarrow+CO_2+H_2O\)

       0,3                          0,3                          ( mol )

\(V_{CO_2}=\left(0,1+0,6\right).22,4=15,68\left(g\right)\)

\(m=0,3.197=59,1\left(g\right)\)

\(m_{NaCl}=90.5\%=4,5\left(g\right)\)

\(m_{dd}=10+90=100\left(g\right)\)

\(\left\{{}\begin{matrix}C\%_{NaOH}=\dfrac{10}{100}.100=10\%\\C\%_{NaCl}=\dfrac{4,5}{100}.100=4,5\%\end{matrix}\right.\)

\(n_{hhk}=\dfrac{17,92}{22,4}=0,8\left(mol\right)\)

\(\dfrac{M_{hhk}}{M_{H_2}}=6,25\Rightarrow M_{hhk}=6,25.2=12,5\) 

Đặt \(\left\{{}\begin{matrix}n_{H_2}=x\\n_{CH_4}=y\end{matrix}\right.\) ( mol ) \(\Rightarrow x+y=0,8\left(1\right)\)

\(\Rightarrow M_{hhk}=\dfrac{2x+16y}{x+y}=12,5\) \(\Leftrightarrow\dfrac{2x+16y}{0,8}=12,5\) \(\Leftrightarrow x+8y=5\) (2)

\(\left(1\right);\left(2\right)\rightarrow\left\{{}\begin{matrix}x=0,2\\y=0,6\end{matrix}\right.\)

\(n_{O_2}=\dfrac{51,2}{32}=1,6\left(mol\right)\)

\(2H_2+O_2\rightarrow\left(t^o\right)2H_2O\)

0,2       0,1               0,2     ( mol )

\(CH_4+2O_2\rightarrow\left(t^o\right)CO_2+2H_2O\)

 0,6       1,2                0,6         1,2  ( mol )

\(n_{O_2\left(p.ứ\right)}=0,1+1,2=1,3< 1,6\) 

\(m_{H_2O}=\left(0,2+1,2\right).18=25,2\left(g\right)\)

hh Y gồm: \(CO_2\) và \(O_{2\left(dư\right)}\)

\(\left\{{}\begin{matrix}\%V_{CO_2}=\dfrac{0,6}{0,6+0,3}.100=66,67\%\\\%V_{O_2\left(dư\right)}=100\%-66,67\%=33,33\%\end{matrix}\right.\)

\(\left\{{}\begin{matrix}\%m_{CO_2}=\dfrac{0,6.44}{0,6.44+0,3.16}.100=84,61\%\\\%m_{O_2\left(dư\right)}=100\%-84,61\%=18,39\%\end{matrix}\right.\)

sửa C: \(=5x\left(y-z\right)^2\)

`a.`\(x\left(x+1\right)^2+x\left(x-5\right)-5\left(x+1\right)^2\)

\(=\left(x+1\right)^2\left(x-5\right)+x\left(x-5\right)\)

\(=\left(x-5\right)\left[\left(x+1\right)^2+x\right]\)

\(=\left(x-5\right)\left(x^2+3x+1\right)\)

`b.`\(3x^2-12y^2=3\left(x^2-4y^2\right)=3\left(x-2y\right)\left(x+2y\right)\)

`c.`\(5xy^2-10xyz+5xz^2=5x\left(y^2-2yz+z^2\right)=5x\left(y+z\right)^2\)

`d.`\(x^3+3x^2+3x+1-27z^3=\left(x+1\right)^3-\left(3z\right)^3=\left(x+1-3z\right)\left[\left(x+1\right)^2+3z\left(x+1\right)+9z^2\right]\)

`a.`\(x^4+x^3+x+1=x^3\left(x+1\right)+\left(x+1\right)=\left(x+1\right)\left(x^3+1\right)=\left(x+1\right)\left(x+1\right)\left(x^2-x+1\right)=\left(x+1\right)^2\left(x^2-x+1\right)\)

`c.`\(x^2y+xy^2-x-y=xy\left(x+y\right)-\left(x+y\right)=\left(x+y\right)\left(xy-1\right)\)

`d.`\(ax^2+ay-bx^2-by=a\left(a^2+y\right)-b\left(x^2+y\right)=\left(x^2+y\right)\left(a-b\right)\)