Đáp án:

Giải thích các bước giải:

 1)

$Cu+2H2SO4--->CuSO4+2H2O+SO2$

$x--------------------x(mol)$

$2Fe+6H2SO4----->Fe2(SO4)3+3SO2+6H2O$

$y-----------------------1,5y(mol)$

Ta có

$nSO2=10,08/22,4=0,45(mol)$

Theo Bài ta có hpt

$64x+56y=20,8$

$x+1,5y=0,45$

=>$x=0,15,y=0,2$

=>%$mCu=0,15.64/20,8.100$%$=46,15$%

=>%$mFe=100-46,15=53,84$%

2)

$2Al+6H2SO4--->Al2(SO4)3+3SO2+6H2O$

$CuO+H2SO4--->CuSO4+H2O$

$nAl=2/3nSO2=0,2(mol)$

=>$mAl=0,2.27=5,4(g)$

=>%$mAl=5,4/7,8.100$%$=69,23$%

=>%$mCuO=100-69,23=30,77$%

$1/$

$n_{SO_2}=10,08/22,4=0,45mol$

$Cu+2H_2SO_4\to SO_2+2H_2O+CuSO_4$

$2Fe+6H_2SO_4\to Fe_2(SO_4)_3+6H_2O+3SO_2$

Gọi $n_{Cu}=a;n_{Fe}=b$

Ta có :

$m_{hh}=64a+56b=20,8$

$n_{SO_2}=a+1,5b=0,45$

$\left\{\begin{matrix}
64a+56b=20,8 & \\ 
2a+1,5b=0,45 & 
\end{matrix}\right.$

$⇔\left\{\begin{matrix}
a=0,15 & \\ 
y=0,2 & 
\end{matrix}\right.$

$⇒\%m_{Cu}=\dfrac{0,15.64.100\%}{20.8}=46,153\%$

$⇒\%m_{Fe}=100\%-46,153\%=53,847\%$

2/

$n_{SO_2}=6,72/22,4=0,3mol$

$PTHH :$

$2Al+6H_2SO_4\to Al_2(SO_4)_3+6H_2O+3SO_2(1)$

$CuO+H_2SO_4\to H_2O+CuSO_4(2)$

Theo pt (1) :

$n_{Al}=2/3.n_{SO_2}=2/3.0,3=0,2mol$

$⇒m_{Al}=0,2.27=5,4g$

$⇒\%m_{Al}=\dfrac{5,4.100\%}{7,8}=69,23\%$

$⇒\%m_{CuO}=100\%-69,23\%=30,77\%$