câu 1 
\(n_{Fe}=\dfrac{14}{56}=0,25\left(mol\right)\\ pthh:Fe+2HCl\rightarrow FeCl_2+H_2\) 
           0,25   0,5         0,25       0,25 
\(m_{FeCl_2}=0,25.127=31,75g\\ V_{H_2}=0,25.22,4=5,6\\ C_{M\left(HCl\right)}=\dfrac{0,5}{0,2}=2,5M\)
câu 2 
1 ) \(m_{\text{dd}}=35+100=135g\\ 2,C\%=\dfrac{204}{204+100}.100=60\%\\ =>m\text{dd}=\dfrac{100.204}{60}=340g\)
 

nAl = 5,4/27 = 0,2 (mol)

PTHH: 2Al + 3H2SO4 -> Al2(SO4)3 + 3H2

nH2 = 0,2 : 2 . 3 = 0,3 (mol)

VH2 = 0,3 . 22,4 = 6,72 (l)

PTHH: Fe2O3 + 3H2 -> (t°) 2Fe + 3H2O

nFe = 0,3 : 3 . 2 = 0,2 (mol)

a = mFe = 0,2 . 56 = 11,2 (g)

nFe2O3 = 0,3/3 = 0,1 (mol)

mFe2O3 = 0,1 . 160 = 16 (g)

m = 16/60% = 80/3 (g)

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

Ta có: \(n_{Fe}=\dfrac{0,56}{56}=0,01\left(mol\right)\)

Theo PT: \(n_{H_2SO_4}=n_{FeSO_4}=n_{H_2}=n_{Fe}=0,01\left(mol\right)\)

b, \(m_{FeSO_4}=0,01.152=1,52\left(g\right)\)

\(V_{H_2}=0,01.22,4=0,224\left(l\right)\)

c, \(m_{ddH_2SO_4}=\dfrac{0,01.98}{19,6\%}=5\left(g\right)\)

a: \(Fe+2HCl\rightarrow FeCl_2+H_2\uparrow\)

0,1           0,2       0,1           0,1

b: \(V_{H_2}=0.1\cdot22.4=2.24\left(lít\right)\)

c: \(m_{FeCl_2}=0.1\left(56+35.5\cdot2\right)=12.7\left(g\right)\)

a, PT: \(Fe+H_2SO_4\rightarrow FeSO_4+H_2\)

b, Ta có: \(n_{Fe}=\dfrac{19,6}{56}=0,35\left(mol\right)\)

Theo PT: \(n_{H_2}=n_{Fe}=0,35\left(mol\right)\Rightarrow V_{H_2}=0,35.22,4=7,84\left(l\right)\)

c, \(n_{H_2SO_4}=n_{Fe}=0,35\left(mol\right)\Rightarrow C_{M_{H_2SO_4}}=\dfrac{0,35}{0,2}=1,75\left(M\right)\)

d, \(n_{FeSO_4}=n_{Fe}=0,35\left(mol\right)\Rightarrow m_{FeSO_4}=0,35.152=53,2\left(g\right)\)

e, \(C_{M_{FeSO_4}}=\dfrac{0,35}{0,2}=1,75\left(M\right)\)

d, \(n_{H_2SO_4}=0,25.1,6=0,4\left(mol\right)\)

Xét tỉ lệ: \(\dfrac{n_{Fe}}{1}< \dfrac{n_{H_2SO_4}}{1}\), ta được H₂SO₄ dư.

Theo PT: \(n_{H_2SO_4\left(pư\right)}=n_{Fe}=0,35\left(mol\right)\)

\(\Rightarrow n_{H_2SO_4\left(dư\right)}=0,4-0,35=0,05\left(mol\right)\)

\(\Rightarrow m_{H_2SO_4\left(dư\right)}=0,05.98=4,9\left(g\right)\)

giup mink voi !!!

gap ah

 \(n_{Fe}=0,1\left(mol\right);n_{Cl_2}=0,12\left(mol\right)\)

\(2Fe+3Cl_2\rightarrow2FeCl_3\) (1)

Lập tỉ lệ : \(\dfrac{0,1}{2}>\dfrac{0,12}{3}\)

=> Sau phản ứng Fe dư

Hòa tan sản phẩm vào nước ta được:

2FeCl3 + Fe → 3FeCl2 (2)

b) Muối tạo thành là FeCl2 và FeCl3

\(n_{Fe\left(dư\right)}=0,1-0,12.\dfrac{2}{3}=0,02\left(mol\right)\)

\(n_{FeCl_2}=3n_{Fe\left(dư\right)}=0,06\left(mol\right)\)

Ta có : \(n_{FeCl_3}=n_{FeCl_3\left(1\right)}-n_{FeCl_3\left(2\right)}=0,12.\dfrac{2}{3}-0,02.2=0,04\left(mol\right)\)

\(m_{FeCl_2}=0,06.127=7,62\left(g\right)\)

\(m_{FeCl_3}=0,04.162,5=6,5\left(g\right)\)

c) \(CM_{FeCl_2}=\dfrac{0,06}{0,5}=0,12M\)

\(CM_{FeCl_3}=\dfrac{0,04}{0,5}=0,08M\)

a)\(n_{Fe}=\dfrac{14}{56}=0,25\left(mol\right)\)

PTHH: Fe + 2HCl → FeCl₂ + H₂

Mol:    0,25    0,5                    0,25

b) \(C_{M_{ddHCl}}=\dfrac{0,5}{0,5}=1M\)

c) \(V_{H_2}=0,25.22,4=5,6\left(l\right)\)

a) Fe+2HCl --> FeCl₂+H₂  

b) tự tính

c) tự tính

a, Fe+2HCl=>FeCl₂+H₂

_b, N_Fe=\(\frac{14}{56}\)=0,25mol     N_HCl=\(\frac{0,2}{22,4}\)\(\approx\)0,0089mol

ta có \(\frac{0,25}{1}< \frac{0,0089}{1}\left(Fe.hết.HCl.dư\right)\)

=>N_FeCl2=0,25mol

=>M_FeCL2=0,25x127=31,75g

c,N_H2=0,25mol

  v_h2=0,25x22,4=5,6mol

d, mình chưa học phần này 

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

PT: \(Fe+2HCl\rightarrow FeCl_2+H_2\)

Theo PT: \(n_{H_2}=n_{FeCl_2}=n_{Fe}=0,2\left(mol\right)\)

\(\Rightarrow V_{H_2}=0,2.22,4=4,48\left(l\right)\)

b, \(m_{FeCl_2}=0,2.127=25,4\left(g\right)\)

c, \(n_{HCl}=2n_{H_2}=0,4\left(mol\right)\Rightarrow C_{M_{HCl}}=\dfrac{0,4}{0,3}=\dfrac{4}{3}\left(M\right)\)

\(C_{M_{FeCl_2}}=\dfrac{0,2}{0,3}=\dfrac{2}{3}\left(M\right)\)