1,

$n_{SO_2}=\dfrac{3,36}{22,4}=0,15\ (mol)$

$\to n_{SO_4^{2-}\ (TM)}=n_{SO_2}=0,15\ (mol)$

$\to m_{muối}=m_{hh}+m_{SO_4^{2-}}=5+0,15.96=19,4\ (g)$

2,

$n_{Fe}=\dfrac{3,36}{56}=0,06\ (mol)$

$\to n_{FeSO_4}=n_{Fe}=0,06\ (mol)$

$6FeSO_4+K_2CrO_7+7H_2SO_4\to 3Fe_2(SO_4)_3+Cr_2(SO_4)_3+K_2SO_4+7H_2O$

$n_{K_2CrO_7}=\dfrac16n_{FeSO_4}=0,01\ (mol)$

$\to V_{K_2CrO_7}=\dfrac{0,01}{1}=0,01\ (l)=10\ (ml)$

 

Đáp án:

 `1. m=19,4(g)`

`2. V=10ml`

Giải thích các bước giải:

 1.

Sơ đồ :

\(\rm\underbrace{\begin{cases}\mathop{Fe}\limits^{0}\\\mathop{Mg}\limits^{0}\\\mathop{Cu}\limits^{0}\\\end{cases}}_{5(g)}\xrightarrow{+H_2\mathop{S}\limits^{6}O_4 đ,t^o}\left\langle\begin{array}{l}\rm\mathop{S}\limits^{+4}O_2-0,15\\\begin{cases}Fe^{3+}\\Mg^{2+}\\Cu^{2+}\\SO_4^{2-}\\\end{cases}\end{array} \right.\)

$\xrightarrow{BTe}2n_{SO_2}=3n_{Fe}+2n_{Mg}+2n_{Cu}$

$\xrightarrow{BTĐT}2n_{SO_4^{2-}}=3n_{Fe}+2n_{Mg}+2n_{Cu}$

$\to n_{SO_2}=n_{SO_4^{2-}}=0,15(mol)$

`->m=m_{\text{Muối}}=m_{\text{Kim loại}}+m_{SO_4^{2-}}=5+0,15.96=19,4(g)`

2.

`n_{Fe}=\frac{3,36}{56}=0,06(mol)`

`Fe+H_2SO_4->FeSO_4+H_2`

`6FeSO_4+7H_2SO_4+K_2Cr_2O_7->3Fe_2(SO_4)_3+Cr_2(SO_4)_3+K_2SO_4+7H_2O`

Theo phương trình 

`n_{K_2Cr_2O_7}=1/6 n_{Fe}=0,01(mol)`

`->V=\frac{0,01}{1}=0,01(l)=10ml`