Ta có: \(n_{H_2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\)

PT: \(Fe+2HCl\rightarrow FeCl_2+H_2\)

Theo PT: \(n_{HCl}=2n_{H_2}=0,2\left(mol\right)\Rightarrow C_{M_{HCl}}=\dfrac{0,2}{0,1}=2\left(M\right)\)

\(n_{Fe}=n_{H_2}=0,1\left(mol\right)\Rightarrow m_{Fe}=0,1.56=5,6\left(g\right)\)

\(\Rightarrow m_{Cu}=20-5,6=14,4\left(g\right)\)

\(\Rightarrow\left\{{}\begin{matrix}\%m_{Fe}=\dfrac{5,6}{20}.100\%=28\%\\\%m_{Cu}=72\%\end{matrix}\right.\)

a) \(n_{Zn}=\dfrac{13}{65}=0,2\left(mol\right)\)

PTHH: Zn + 2HCl --> ZnCl₂ + H₂

             0,2--->0,4--->0,2--->0,2

=> V = 0,2.22,4 = 4,48 (l)

b) \(C_{M\left(dd.HCl\right)}=\dfrac{0,4}{0,5}=0,8M\)

c) \(m_{ZnCl_2}=0,2.136=27,2\left(g\right)\)

a) $Fe + 2HCl \to FeCl_2 + H_2$

b) $n_{HCl} = 0,2.0,3 = 0,06(mol)$

Theo PTHH : 

$n_{Fe} = n_{H_2} = \dfrac{1}{2}n_{HCl} = 0,03(mol)$

Suy ra : 

$m = 0,03.56 = 1,68(gam)$
$V = 0,03.22,4 = 0,672(lít)$

b) $n_{FeCl_2} = \dfrac{1}{2}n_{HCl} = 0,03$

$C_{M_{FeCl_2}} = \dfrac{0,03}{0,2} = 0,15M$

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

0,3     0,3             0,3           0,3 

\(n_{H_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\)

\(m_{Fe}=0,3.56=16,8\left(g\right)\)

\(C_{M_{H_2SO_4}}=\dfrac{n}{V}=\dfrac{0,3}{0,2}=1,5M\)

\(C_{M_{FeSO_4}}=\dfrac{n}{V}=\dfrac{0,3}{0,2}=1,5M\)

PTHH: Fe + H₂SO₄ --> FeSO₄ + H₂

200ml = 0,2 lít.

a) Số mol H₂: n_H₂ = 6,72 ÷ 22,4 = 0,3 mol

Theo PTHH => Số mol Fe: n_Fe = 0,3 mol

     => Khối lượng Fe: m_Fe = 16,8g

b) Số mol H₂SO₄: n_H₂SO₄ = 0,3 mol

Nồng độ mol dd: C_M = 0,3 ÷ 0,2 = 1,5M

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

PT: \(Fe+2HCl\rightarrow FeCl_2+H_2\)

a, \(n_{H_2}=n_{Fe}=0,2\left(mol\right)\Rightarrow V_{H_2}=0,2.22,4=4,48\left(l\right)\)

b, \(n_{HCl}=2n_{Fe}=0,4\left(mol\right)\Rightarrow C_{M_{HCl}}=\dfrac{0,4}{0,15}=\dfrac{8}{3}\left(M\right)\)

c, \(n_{FeCl_2}=n_{Fe}=0,2\left(mol\right)\Rightarrow C_{M_{FeCl_2}}=\dfrac{0,2}{0,15}=\dfrac{4}{3}\left(M\right)\)

$a\big)$

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

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

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

Fe+2HCl->FeCl2+H2

0,2---0,4---0,2----0,2 mol

nFe=11,2\56=0,2 mol

=>VH2=0,2.22,4=4,48l

=>CMHCl=0,2\0,5=0,4 M

a, PTPƯ: SO₃ + H₂O ---> H₂SO₄

n_SO3=\(\dfrac{2,24}{22,4}=0,1mol\)

1 mol SO₃ ---> 0,1 mol H₂SO₄

nên 0,1 mol SO₃ ---> 0,1 mol H₂SO₄

C_{M H2SO4}=\(\dfrac{0,1}{0,5}\)=0,2 M

b, PTPƯ: Zn + H₂SO₄ ---> ZnSO₄ + H₂

1 mol H₂SO₄ ---> 1 mol Zn

nên 0,1 mol H₂SO₄ ---> 0,1 mol Zn

m_Zn=0,1.65=6,5 g

Giải hộ mình vs ạaaa