\(n_{H2}=\frac{2,688}{22,4}=0,12\left(mol\right)\)
\(2A+2nHCl\rightarrow2ACln+nH_2\)
0,24/n____________0,12________
Ta có \(A.\frac{0,24}{n}=7,8\)
\(\Rightarrow A=\frac{65n}{2}\)
\(\Rightarrow\left\{{}\begin{matrix}n=2\\A=65\end{matrix}\right.\)
\(\Rightarrow\) A là kim loại Zn
\(n_{HCl}=2n_{H2}=2.0,12=0,24\)
Gọi oxit là B2On
\(PTHH:B_2O_n+2nHCl\rightarrow2BCl_n+nH_2O\)
______0,12/n______0,24_____________________
\(\frac{\left(2B+16n\right).0,12}{n}=6,4\)
\(\Rightarrow B=\frac{56n}{3}\)
\(\Rightarrow\left\{{}\begin{matrix}n=3\\B=56\end{matrix}\right.\)
\(\Rightarrow\) B là Fe sắt
P/s : Xác định A và B nhé
2A+2xHCl--->2AClx+xH2
n H2=2,688/22,4=0,12(mol)
n A=2/x n H2=0,24/x(mol)
M A=7,8 : 0,24/x =32,5x
+x=2--->M A=65(Zn)
n HCl=2n H2=0,24(mol)
\(B_xO_y+2yHCl\rightarrow xBCl_{\frac{2y}{x}}+yH2O\)
n BxOy=1/2y n HCl=0,12y(mol)
M BxOy= 6,4/0,12y=53,333
y=3-->M=160
xB+ 16.3=160
-->xB=112
x=2--->B=56(Fe)
Vậy A là Zn,B là Fe