Câu 38:

Gọi $x$, $y$ là số mol $Al$, $Fe$

$n_{Cl_2}=\dfrac{20,16}{22,4}=0,9(mol)$

Bảo toàn e: $3n_{Al}+3n_{Fe}=2n_{Cl_2}$

$\Rightarrow 3x+3y=0,9.2=1,8$        $(1)$

$n_{H_2}=\dfrac{15,68}{22,4}=0,7(mol)$

Bảo toàn e: $3n_{Al}+2n_{Fe}=2n_{H_2}$

$\Rightarrow 3x+2y=0,7.2=1,4$        $(2)$

$(1)(2)\Rightarrow x=0,2; y=0,4$

$\to a=0,2.27+0,4.56=27,8g$

$n_{HCl}=2n_{H_2}=1,4(mol)$

$\to m_{dd HCl}=1,4.36,5:25\%=204,4g$

Câu 40: 

a,

$n_{H_2}=\dfrac{0,672}{22,4}=0,03(mol)$

$\Delta m_{dd}=4,05g=m_X-m_{H_2}$

$\Rightarrow m=m_X=4,05+0,03.2=4,11g$

Gọi $n$ là hoá trị của $X$

$2X+2nHCl\to 2XCl_n+nH_2$

$\Rightarrow n_X=\dfrac{0,06}{n}(mol)$

$M_X=\dfrac{4,11n}{0,06}=68,5n$

$\to n=2; M_X=137(Ba)$

Vậy kim loại $X$ là bari.

b,

$n_{HCl\text{pứ}}=2n_{H_2}=0,06(mol)$

$n_{NaOH}=\dfrac{25.16\%}{40}=0,1(mol)$

$NaOH+HCl\to NaCl+H_2O$

$\Rightarrow n_{HCl\text{dư}}=0,1(mol)$

$C_{M_{HCl}}=\dfrac{0,1+0,06}{0,05}=3,2M$