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,
$n_{HCl\text{pứ}}=2n_{H_2}=0,06(mol)$
$n_{NaOH}=\dfrac{25.16\%}{40}=0,1(mol)$
$NaOH+HCl\to NaCl+H_2O$
$\Rightarrow n_{HCl\text{dư}}=0,1(mol)$
$C_{M_{HCl}}=\dfrac{0,1+0,06}{0,05}=3,2M$
