\(n_{H_2SO_4}=0,25.2=0,5\left(mol\right)\)

Gọi \(\left\{{}\begin{matrix}n_{Al_2O_3}=x\left(mol\right)\\n_{CuO}=y\left(mol\right)\end{matrix}\right.\)

\(Al_2O_3+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2O\)

 x----------> 3x --------> x 

\(CuO+H_2SO_4\rightarrow CuSO_4+H_2O\)

y --------> y -------->  y

Có hệ phương trình

\(\left\{{}\begin{matrix}102x+80y=26,2\\3x+y=0,5\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0,1\\y=0,2\end{matrix}\right.\)

\(\%_{m_{Al_2O_3}}=\dfrac{102.0,1.100}{26,2}=38,93\%\)

\(\%_{m_{CuO}}=\dfrac{80.0,2.100}{26,2}=61,07\%\)

\(CM_{Al_2\left(SO_4\right)_3}=\dfrac{x}{0,25}=\dfrac{0,1}{0,25}=0,4M\)

\(CM_{CuSO_4}=\dfrac{y}{0,25}=\dfrac{0,2}{0,25}=0,8M\)

\(n_{H_2SO_4}=0,25.2=0,5mol\\ n_{Al_2O_3}=a,n_{CuO}=b\\ Al_2O_3+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2O\\ CuO+H_2SO_4\rightarrow CuSO_4+H_2O\\ \Rightarrow\left\{{}\begin{matrix}3a+b=0,5\\102a+80b=26,2\end{matrix}\right.\\ \Rightarrow a=0,1;b=0,2\\ \%m_{Al_2O_3}=\dfrac{0,1.102}{26,2}\cdot100=39\%\\ \%m_{CuO}=100-39=61\%\)

Al₂O₃ + 3H₂SO₄\(\rightarrow\)Al₂(SO₄)₃ + 3H₂O (1)

CuO + H₂SO₄\(\rightarrow\)CuSO₄ + H₂O (2)

Đặt n_Al2O3=a

n_CuO=b

Ta có hệ pt:

\(\left\{{}\begin{matrix}102a+80b=26,2\\3a+b=0,5\end{matrix}\right.\)

a=0,1;b=0,2

m_CuO=0,2.80=16(g)

% CuO=\(\dfrac{16}{26,2}.100\%=61\%\)

%Al₂O₃=100-61=39%

giúp mk với!!!

\(n_{FeO}=a\left(mol\right),n_{CuO}=b\left(mol\right)\)

\(m_{hh}=72a+80b=19.2\left(g\right)\left(1\right)\)

\(FeO+H_2SO_4\rightarrow FeSO_4+H_2O\)

\(CuO+H_2SO_4\rightarrow CuSO_4+H_2O\)

\(n_{H_2SO_4}=a+b=0.25\left(mol\right)\left(2\right)\)

\(\left(1\right),\left(2\right):a=0.1,b=0.15\)

\(m_{FeO}=0.1\cdot72=7.2\left(g\right)\)

\(m_{CuO}=12\left(g\right)\)

\(C_{M_{FeSO_4}}=\dfrac{0.1}{0.25}=0.4\left(M\right)\)

\(C_{M_{CuSO_4}}=\dfrac{0.15}{0.25}=0.6\left(M\right)\)

Gọi x,y lần lượt là số mol Al2O3, CuO

Al2O3 + 3H2SO4 → Al2(SO4)3 + 3H2O

CuO + H2SO4 → H2O + CuSO4

\(\left\{{}\begin{matrix}102x+80y=12,3\\3x+y=\dfrac{100.24,5\%}{98}=0,25\end{matrix}\right.\)

=> x= 0,056; y=0,082

=> \(\%m_{Al_2O_3}=\dfrac{0,056.102}{12,3}.100=46,44\%\)

=> %mCuO= 100 - 46,44= 53,56%

b) 

Al2O3 + 6HCl → 2AlCl3 + 3H2O

CuO + 2HCl → CuCl2 + H2O 

=> \(m_{HCl}=\dfrac{(0,056.6+0,082.2).36,5}{7\%}=260,7\%\)