\(n_{CuO}=\dfrac{8}{80}=0,1(mol)\\ PTHH:CuO+2HCl\to CuCl_2+H_2O\\ \Rightarrow n_{HCl}=0,2(mol)\\ \Rightarrow V_{dd_{HCl}}=\dfrac{0,2}{2}=0,1(l)\)

A

B

a

b,\(n_{HCl}=0,2.2=0,4\left(mol\right)\)

PTHH: CuO + 2HCl → CuCl₂ + H₂

Mol:     0,2         0,4  

\(\Rightarrow m_{CuO}=0,2.80=16\left(g\right)\)

c,\(n_{ZnO}=\dfrac{16,2}{81}=0,2\left(mol\right)\)

PTHH: ZnO + H₂SO₄ → ZnSO₄ + H₂

Mol:      0,2         0,2

\(\Rightarrow V_{ddH_2SO_4}=\dfrac{0,2}{2}=0,1\left(l\right)\)

d,\(n_{H_2SO_4}=2.0,1=0,2\left(mol\right)\)

PTHH: H₂SO₄ + 2KOH → K₂SO₄ + 2H₂O

Mol:     0,2           0,4

\(\Rightarrow V_{ddKOH}=\dfrac{0,4}{1}=0,4\left(l\right)\)

Đáp án : A

Đáp án A

Chọn C

Đáp án C

Áp dụng bảo toàn nguyên tố ta có:

Thể tích dung dịch HCl 1M là:

a) \(n_{H_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\)

PTHH: 2Al + 6HCl --> 2AlCl₃ + 3H₂

_____0,2<---0,6<--------------0,3

=> \(\left\{{}\begin{matrix}\%Al=\dfrac{0,2.27}{15,6}.100\%=34,615\%\\\%Al_2O_3=\dfrac{15,6-0,2.27}{15,6}.100\%=65,385\%\end{matrix}\right.\)

b) \(n_{Al_2O_3}=\dfrac{15,6-0,2.27}{102}=0,1\left(mol\right)\)

PTHH: Al₂O₃ + 6HCl --> 2AlCl₃ + 3H₂O

______0,1--->0,6

=> n_HCl = 0,6+0,6 = 1,2(mol)

=> \(V_{dd}=\dfrac{1,2}{2}=0,6\left(l\right)\)