$a\big)$

$M_A=9,4.2=18,8(g/mol)$

$\to \dfrac{n_{CO_2}}{n_{H_2}}=\dfrac{18,8-2}{44-18,8}=\dfrac{2}{3}$

Mà $n_{CO_2}+n_{H_2}=\dfrac{11,2}{22,4}=0,5(mol)$

\(\begin{array} {l} \to n_{CO_2}=0,2(mol);n_{H_2}=0,3(mol)\\ Fe+2HCl\to FeCl_2+H_2\\ FeCO_3+2HCl\to FeCl_2+CO_2+H_2O\\ \text{Theo PT: }n_{Fe}=n_{H_2}=0,3(mol);n_{FeCO_3}=n_{CO_2}=0,2(mol)\\ \to m=0,3.56+0,2.116=40(g) \end{array}\)

$b\big)$

Đổi $400ml=0,4l$

\(\begin{array} {l} \text{Theo PT: }n_{FeCl_2}=n_{H_2}+n_{CO_2}=0,5(mol)\\ \to C_{M\,FeCl_2}=\dfrac{0,5}{0,4}=1,25M \end{array}\)

$c\big)$

\(\begin{array}{l} m_{dd\,FeCl_2}=\dfrac{400}{1,2}\approx 333,33(g)\\ \to C\%_{FeCl_2}=\dfrac{0,5.127}{333,33}.100\%=19,05\%\end{array}\)

 Mời các chiến thần trả lời giúp mình

PTHH: \(Fe+H_2SO_4\rightarrow FeSO_4+H_2\uparrow\)

Ta có: \(\left\{{}\begin{matrix}n_{Fe}=\dfrac{33,6}{56}=0,6\left(mol\right)\\n_{H_2SO_4}=\dfrac{784\cdot10\%}{98}=0,8\left(mol\right)\end{matrix}\right.\)

Xét tỉ lệ: \(\dfrac{0,6}{1}< \dfrac{0,8}{1}\) \(\Rightarrow\) H₂SO₄ còn dư, Fe phản ứng hết

\(\Rightarrow\left\{{}\begin{matrix}n_{FeSO_4}=n_{H_2}=0,6mol\\n_{H_2SO_4\left(dư\right)}=0,2mol\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}m_{FeSO_4}=0,6\cdot152=91,2\left(g\right)\\m_{H_2SO_4\left(dư\right)}=0,2\cdot98=19,6\left(g\right)\\m_{H_2}=0,6\cdot2=1,2\left(g\right)\end{matrix}\right.\)

Mặt khác: \(m_{dd}=m_{Fe}+m_{ddH_2SO_4}-m_{H_2}=816,4\left(g\right)\)

\(\Rightarrow\left\{{}\begin{matrix}C\%_{FeSO_4}=\dfrac{91,2}{816,4}\cdot100\%\approx11,17\%\\C\%_{H_2SO_4\left(dư\right)}=\dfrac{19,6}{816,4}\cdot100\%\approx2,4\%\end{matrix}\right.\)

\(n_{Fe}=\dfrac{33,6}{56}=0,6\left(mol\right)\)

\(n_{H2SO4}=\dfrac{784.10\%}{98}=0,8\left(mol\right)\)

PTHH : \(Fe+H_2SO_4-->FeSO_4+H_2\uparrow\)

Theo pthh : \(n_{H2}=n_{FeSO4}=n_{H2SO4\left(pứ\right)}=n_{Fe}=0,6\left(mol\right)\)

\(\Rightarrow n_{H2SO4\left(dư\right)}=0,8-0,6=0,2\left(mol\right)\)

Áp dụng ĐLBTKL :

m_Fe + m_{(dd H2SO4)} = m_(ddspu) + m_H2

=> 33,6 + 784 = m_(ddspu) + 0,6.2

=> m_(ddspu) = 816,4(g)

\(\Rightarrow\left\{{}\begin{matrix}C\%FeSO_{\text{4}}=\dfrac{0,6.152}{816,4}\cdot100\%\approx11,17\%\\C\%H_2SO_{4\left(dư\right)}=\dfrac{0,2.98}{816,4}\cdot100\%\approx2,4\%\end{matrix}\right.\)

n_Fe=33,6 : 56 =0,6 (mol)

m_H2SO4=784.10%=78,4(g)

n_H2SO4=78,4:98=0,8(mol)

             PTHH:

Fe + H₂SO₄ —> FeSO₄ + H₂O

Từ PTHH ta suy ra : Fe hết , H₂SO₄ dư 

=>n_FeSO4=n_{H2SO4 p/ứ}=n_Fe=0,6(mol)

=>n_{H2SO4 dư}=0,8-0,6=0,2(mol)

m_{dd sau p/ứ}=784+33,6=817,6(g)

%m_{H2SO4 dư}=\(\dfrac{0,2.98}{817,6}.100\%\approx2,4\%\)

%m_FeSO4=\(\dfrac{0,6.152}{817,6}.100\%\approx11,2\%\)

Chúc bạn học tốt UwU

`Fe + H_2 SO_4 -> FeSO_4 + H_2`

`0,25`     `0,25`                        `0,25`       `(mol)`

`a)n_[Fe]=[22,4]/56=0,4(mol)`

`n_[H_2 SO_4]=[24,5]/98=0,25(mol)`

Có: `[0,4]/1 > [0,25]/1=>Fe` hết, `H_2 SO_4` 

  `=>m_[Fe(dư)]=(0,4-0,25).56=8,4(g)`

`b)V_[H_2]=0,25.22,4=5,6(l)`

   Ko được ghi `Fe+H_2 SO_4->Fe_2 (SO_4)_3+H_2` vì đây là `H_2 SO_4` loãng

\(n_{Ba}=\dfrac{6,85}{137}=0,05\left(mol\right)\\ m_{H_2SO_4}=500.1,96\%=9,8\left(g\right)\\ PTHH:Ba+H_2SO_4\rightarrow BaSO_4+H_2\uparrow\\ LTL:0,05< 0,1\Rightarrow H_2SO_4.dư\)

\(n_{BaSO_4}=n_{H_2SO_4\left(pư\right)}=n_{Ba}=n_{H_2}=0,05\left(mol\right)\)

\(\Rightarrow n_{H_2SO_4\left(dư\right)}=0,1-0,05=0,05\left(mol\right)\)

\(V_{dd}=\dfrac{500}{1,15}\approx434\left(ml\right)=0,434\left(l\right)\)

\(C_{MBaSO_4}=\dfrac{0,05}{0,434}=0,115M\\ C_{MH_2SO_4\left(dư\right)}=\dfrac{0,05}{0,434}=0,115M\)

a) Số mol của 11,2 gam Fe:

\(n_{Fe}=\dfrac{11,2}{56}=0,2\left(mol\right)\)

PTHH:

\(Fe+H_2SO_4\rightarrow FeSO_4+H_2\uparrow\)

1    :  1         :     1      :      1

0,2-> 0,2    :        0,2   :     0,2  (mol)

Khối lượng của 0,2 mol H2:

\(m_{H_2}=n.M=0,2.2=0,4\left(g\right)\)

b) khồi lượng của 0,2 mol H2SO4:

\(m_{H_2SO_4}=n.M=0,2.98=19,6\left(g\right)\)

Nồng độ phần trăm của dd:

\(C\%_{H_2SO_4}=\dfrac{m_{ct}}{m_{dd}}.100\%=\dfrac{19,6}{200}.100\%=9,8\%\)

,nFe2(SO4)3=0.25*1=0.25mol=>nFe3+=0.5 mol 
spư thu dc ran z ma ko tan trong H2SO4 loang=>ran Z la Cu,nCu dư=3.28/64=0.05125mol=>Fe tan het 
va toan bo Fe(3+) chuyen len het Fe(2+) do Cu dư 
dat nFe=x,nCu pư=y=>56x+64y+3.28=17.8 (1) 
mat khac theo bt e 
Fe=Fe(2+)+2e................Fe(3+)+1e=F... 
x-------------> 2x 0.5---->0.5 
Cu=Cu(2+)+2e 
y------------->2y 
vi ne cho=ne nhan=>2x+2y=0.5(2)=>x=0.185,y=0.065 
mCu=3.28+0.065*64=7.44g 

Đáp án B

\(n_{Fe}=\dfrac{16,8}{56}=0,3\left(mol\right)\)

\(Fe+H_2SO_4\rightarrow FeSO_4+H_2\)

\(0,3->0,3-->0,3->0,3\)

\(mH_2SO_4=0,3.98=29,4\left(g\right)\)

\(\Rightarrow C\%_{ddH_2SO_4}=\dfrac{29,4.100}{200}=14,7\%\)

\(V_{FeSO_4}=\dfrac{n}{CM}=\dfrac{0,3}{2}=0,25\left(l\right)\)

\(VH_2=0,3.22,4=6,72\left(l\right)\)

\(n_{Fe}=\dfrac{16,8}{56}=0,3\left(mol\right)\\ pthh:Fe+H_2SO_4\rightarrow FeSO_4+H_2\)
          0,3        0,3          0,3           0,3 
\(C\%_{H_2SO_4}=\dfrac{0,3.98}{200}.100\%=14,7\%\\ V_{FeSO_4}=\dfrac{0,3}{2}=0,15\left(l\right)\\ V_{H_2}=0,3.22,4=6,72\left(l\right)\)

\(a,n_{Zn}=\dfrac{19,5}{65}=0,3\left(mol\right)\\ m_{H_2SO_4}=200.19,6\%=39,2\left(g\right)\\ \rightarrow n_{H_2SO_4}=\dfrac{39,2}{98}=0,4\left(mol\right)\)

PTHH: \(Zn+H_2SO_4\rightarrow ZnSO_4+H_2\uparrow\)

bđ        0,3    0,4

pư        0,3        0,3

spư      0           0,1           0,3        0,3

\(\rightarrow V_{H_2}=0,3.22,4=6,72\left(l\right)\)

\(b,m_{dd}=19,5+200-0,3.2=218,9\left(g\right)\\ \rightarrow\left\{{}\begin{matrix}C\%_{ZnSO_4}=\dfrac{0,3.161}{218,9}.100\%=22,06\%\\C\%_{H_2SO_4\left(dư\right)}=\dfrac{0,1.98}{218,9}.100\%=4,48\%\end{matrix}\right.\)

Tham Khảo