Theo đề gọi \(\left\{{}\begin{matrix}n_A=x\left(mol\right)\\n_B=x\left(mol\right)\end{matrix}\right.\)
\(n_{HCl}=\dfrac{250.18,25\%}{100\%}:36,5=1,25\left(mol\right)\)
\(n_{H_2}=\dfrac{8,4}{22,4}=0,375\left(mol\right)\)
PTHH:
\(A+2HCl\rightarrow ACl_2+H_2\) (1)
x ---> 2x---------------->x
\(2B+6HCl\rightarrow2BCl_3+3H_2\)(2)
x ---->3x------------------>1,5x
Từ PT (1), (2) có: \(n_{H_2.pứ}=\dfrac{1}{2}n_{HCl.pứ}=\dfrac{1}{2}.1,25=0,625\left(mol\right)>n_{H_2.thoát.ra.theo.đề}\)
\(\Rightarrow\) HCl dư.
\(n_{H_2}=x+1,5x=0,375\left(mol\right)\)
\(\Rightarrow x=0,15\)
Mặt khác:
\(m_{hh}=m_A+m_B=Ax+Bx=x\left(A+B\right)=7,65\left(g\right)\\ \Rightarrow A+B=\dfrac{7,65}{0,15}=51g/mol\)
\(\Rightarrow\left\{{}\begin{matrix}B=27\left(Al\right)\\A=24\left(Mg\right)\end{matrix}\right.\)
Vậy A là kim loại Mg, B là kim loại Al.
