1/

$n_{AlCl_3}=26,7/133,5=0,2mol$

$2Al+3Cl_2\to 2AlCl_3$

a/Theo pt :

$n_{Cl_2}=3/2.n_{Cl_2}=3/2.0,2=0,3mol$

$⇒m_{Al}=0,3.27=8,1g$

 2/

$2Fe+3Cl2\to 2FeCl3$

$2Al+3Cl2\to 2AlCl3$

Gọi $n_{Fe}=x ; n_{Al}=y$

Ta có

$56x+27y=8,3$

$162,5x+133,5y=29,6$

Ta có hpt :

$\left\{\begin{matrix}
56x+27y=8,3 & \\ 
162,5x+133,5y=29,6 & 
\end{matrix}\right.$

$⇔\left\{\begin{matrix}
x=0,1 & \\ 
y=0,1 & 
\end{matrix}\right.$

$⇒\%mFe=\dfrac{56.0,1}{8,3}.100\%=67,47\%$

$\%mAl=100\%-67,47\%=32,53\%$

$n_{Cl_2}=\dfrac{0,1.3}{2}+\dfrac{0,1.3}{2}=0,3mol$

$⇒V_{Cl_2}=0,3.22,4=6,72 l$

Đáp án:

Giải thích các bước giải:

$1/$

$PTPƯ:2Al+3Cl_2\xrightarrow{} 2AlCl_3$

$n_{AlCl_3}=\dfrac{26,7}{133,5}=0,2mol.$

$Theo$ $pt:$ $n_{Cl_2}=\dfrac{3}{2}n_{AlCl_3}=0,23mol.$

$⇒V_{Cl_2}=0,3.22,4=6,72l.$

$2/$

$PTPƯ:$

$2Fe+3Cl_2\xrightarrow{} 2FeCl_3$  $(1)$

$2Al+3Cl_2\xrightarrow{} 2AlCl_3$  $(2)$

Gọi $n_{Fe}$ là a, $n_{Al}$ là b.

Theo đề bài ta có hệ pt:

$\left \{ {{56a+27b=8,3} \atop {162,5a+133,5b=29,6}} \right.$ $⇒\left \{ {{a=0,1} \atop {b=0,1}} \right.$ 

$⇒\%m_{Fe}=\dfrac{56.0,1}{8,3}.100\%=67,5\%$

$⇒\%m_{Al}=100\%-67,5\%=32,5\%$

$⇒n_{Cl_2}=\dfrac{0,1.3}{2}+\dfrac{0,1.3}{2}=0,3mol.$

$⇒V_{Cl_2}=0,3.22,4=6,72l.$

chúc bạn học tốt!