\(n_{BaSO_4}=\dfrac{32,62}{233}=0,14\left(mol\right)\)
Gọi \(\left\{{}\begin{matrix}n_{Fe}=a\left(mol\right)\\n_{Cu}=b\left(mol\right)\end{matrix}\right.\)
PTHH:
\(H_2SO_4+BaCl_2\rightarrow BaSO_4\downarrow+2HCl\)
0,14<------------------0,14
\(SO_2+2H_2O+Br_2\rightarrow H_2SO_4+2HBr\uparrow\)
0,14<------------------------0,14
\(2Fe+6H_2SO_4\rightarrow Fe_2\left(SO_4\right)_3+3SO_2\uparrow+6H_2O\)
a-------------------------------------->1,5a
\(Cu+2H_2SO_4\rightarrow CuSO_4+SO_2\uparrow+2H_2O\)
b----------------------------------->b
=> \(\left\{{}\begin{matrix}56a+64b=7,36\\1,5a+b=0,14\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=0,04\left(mol\right)\\b=0,08\left(mol\right)\end{matrix}\right.\)
\(\rightarrow\left\{{}\begin{matrix}\%m_{Fe}=\dfrac{0,04.56}{7,36}.100\%=30,43\%\\\%m_{Cu}=100\%-30,43\%=69,57\%\end{matrix}\right.\)
Gọi \(\left\{{}\begin{matrix}n_{Fe}=x\\n_{Cu}=y\end{matrix}\right.\) \(\rightarrow56x+64y=7,36\) ( 1 )
\(2Fe+6H_2SO_4\left(đ\right)\rightarrow\left(t^o\right)Fe_2\left(SO_4\right)_3+3SO_2+6H_2O\)
x 3x 1/2 x 3/2 x ( mol )
\(Cu+2H_2SO_4\left(đ\right)\rightarrow\left(t^o\right)CuSO_4+SO_2+2H_2O\)
y 2y y y ( mol )
\(\rightarrow\) \(n_{SO_2}=\dfrac{3}{2}x+y\) ( mol )
\(SO_2+Br_2+2H_2O\rightarrow2HBr+H_2SO_4\)
\(\dfrac{3}{2}x+y\) \(\dfrac{3}{2}x+y\) ( mol )
\(n_{BaSO_4}=\dfrac{32,62}{233}=0,14mol\)
\(BaCl_2+H_2SO_4\rightarrow BaSO_4\downarrow+2HCl\)
\(\dfrac{3}{2}x+y\) \(\dfrac{3}{2}x+y\) ( mol )
\(\rightarrow\dfrac{3}{2}x+y=0,14\) ( 2 )
Từ (1) và (2) \(\rightarrow\left\{{}\begin{matrix}x=0,04\\y=0,08\end{matrix}\right.\)
\(\rightarrow\left\{{}\begin{matrix}\%m_{Fe}=\dfrac{0,04.56}{7,36}.100=30,43\%\\\%m_{Cu}=100\%-30,43\%=69,57\%\end{matrix}\right.\)
