a,
$2R+2nH_2SO_4\to R_2(SO_4)_n+nSO_2+2nH_2O$
$n_R=\dfrac{11,2}{M_R}(mol)$
$\Rightarrow n_{SO_2}=\dfrac{5,6n}{M_R}(mol)$
$n_{NaOH}=0,35.2=0,7(mol)$
* Nếu chỉ tạo muối:
Gọi $a$, $b$ là số mol $NaHSO_3$, $Na_2SO_3$
$\Rightarrow 104a+126b=41,8$
$NaOH+SO_2\to NaHSO_3$
$2NaOH+SO_2\to Na_2SO_3+H_2O$
$\Rightarrow a+2b=0,7$
Hệ nghiệm âm, loại.
* Nếu còn dư $NaOH$:
Gọi $x$ là số mol $SO_2$
$2NaOH+SO_2\to Na_2SO_3+H_2O$
$\Rightarrow n_{Na_2SO_3}=x (mol); n_{NaOH}=0,7-2x (mol)$
$\Rightarrow 126x+40(0,7-2x)=41,8$
$\Leftrightarrow x=0,3$
$\dfrac{5,6n}{M_R}=0,3$
$\Leftrightarrow M_R=\dfrac{56n}{3}$
$\to n=3; M_R=56(Fe)$
b,
$n_{Fe}=\dfrac{8,4}{56}=0,15(mol)$
$n_{HCl}=0,2.2=0,4(mol)$
$Fe+2HCl\to FeCl_2+H_2$
$\Rightarrow HCl$ dư
$n_{FeCl_2}=0,15(mol); n_{HCl\text{dư}}=0,4-0,15.2=0,1(mol)$
$AgNO_3+HCl\to AgCl+HNO_3$
$FeCl_2+3AgNO_3\to Fe(NO_3)_3+Ag+2AgCl$
$\Rightarrow n_{Ag}=0,15(mol); n_{AgCl}=0,15.2+0,1=0,4(mol)$
$\to m_{\downarrow}=0,15.108+0,4.143,5= 73,6g$