nH2 = 0,672/22,4 = 0,03 (mol)

PTHH: Fe + 2HCl -> FeCl2 + H2

nFe = 0,03 (mol)

mFe = 0,03 . 56 = 1,68 (g)

mCu = 2,96 - 1,68 = 1,28 (g)

nCu = 1,28/64 = 0,02 (mol)

PTHH:

2Fe + 3Cl2 -> (t°) 2FeCl3

0,03 ---> 0,045 

Cu + Cl2 -> (t°) CuCl2

0,02 ---> 0,02

nCl2 = 0,045 + 0,02 = 0,065 (mol)

VCl2 = 0,065 . 22,4 = 1,456 (l)

\(n_{H2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\)

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

  \(Al_2O_3+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2O\)

b) \(n_{Fe}=n_{H2}=n_{H2SO4}=0,1\left(mol\right)\Rightarrow m_{Fe}=0,1.56=5,6\left(g\right)\)

\(\Rightarrow m_{Al2O3}=15,8-5,6=10,2\left(g\right)\)

c) Ta có : \(n_{Al2O3}=\dfrac{10,2}{102}=0,1\left(mol\right)\Rightarrow n_{H2SO4}=3n_{Al2O3}=0,3\left(mol\right)\)

\(C_{MddH2SO4}=\dfrac{0,1+0,3}{0,2}=2M\)

a)Fe + 2HCl ->FeCl2 + H2\(\uparrow\)

   0.01                                  0.01

FeS + 2HCl ->FeCl2 + H2S\(\uparrow\)

 0.1                                    0.1

H2S + Pb(NO3)2->PbS \(\downarrow\) + 2HNO3

 0.1                             0.1

nPbS =2.39/239=0.1 mol   ,  n (hỗn hợp khí) =2.464/22.4=0.11 mol

n(H2)+n(H2S)=0.11  ->n(H2)=0.01 mol

V(H2)=n * 22.4 = 0.01*22.4=0.224(l)

V(H2S)=n*22.4=0.1*22.4=2.24(l)

m(Fe)=n*M=0.01*56=0.56(g)

m(FeS)=n*M=0.1*88=8.8(g)

Sửa đề: 6,4 gam hh \(\rightarrow\) 6,45 gam hh

a) Đặt \(\left\{{}\begin{matrix}n_{Mg}=a\left(mol\right)\\n_{Al}=b\left(mol\right)\end{matrix}\right.\) \(\Rightarrow24a+27b=6,45\)  (1)

Ta có: \(n_{H_2}=\dfrac{7,28}{22,4}=0,325\left(mol\right)\)

Bảo toàn electron: \(2a+3b=0,65\)  (2)

Từ (1) và (2) \(\Rightarrow\left\{{}\begin{matrix}a=n_{Mg}=0,1\left(mol\right)\\b=n_{Al}=0,15\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}\%m_{Mg}=\dfrac{0,1\cdot24}{6,4}\cdot100\%=37,5\%\\\%m_{Mg}=62,5\%\end{matrix}\right.\)

b) Ta thấy với 6,45 gam hh thì có 0,1 mol Mg và 0,15 mol Al

\(\Rightarrow\) Trong 12,9 gam hh thì chứa 0,2 mol Mg và 0,3 mol Al

Gọi \(n_{SO_2}=x\left(mol\right)\)

Bảo toàn electron: \(2\cdot0,2+3\cdot0,3=2x\) \(\Rightarrow x=n_{SO_2}=0,65\left(mol\right)\)

\(\Rightarrow V_{SO_2}=0,65\cdot22,4=14,56\left(l\right)\)

gọi số mol của Mg là x mol ; Al là y mol => 24x + 27y =6,4

n khí = 7,28/22,4=0,325 mol

bảo toàn e ta có

Mg + H2SO4 --> MgSO4 + H2

 x                                           x           mol

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

  y                                                3/2 y             mol

=> x + 3/2y=0,325

=> x=11/120 mol ; y=7/45 mol

=> mMg11/120*24=2,2g => %mMg = 2,2*100/6,4=34,375%

=>%mAl=100-34,375=65,625%                

Sửa đề: 3,785 (l) → 3,7185 (l)

a, \(2Al+6HCl\rightarrow2AlCl_3+3H_2\)

\(Al_2O_3+6HCl\rightarrow2AlCl_3+3H_2O\)

b, Ta có: \(n_{H_2}=\dfrac{3,7185}{24,79}=0,15\left(mol\right)\)

Theo PT: \(n_{Al}=\dfrac{2}{3}n_{H_2}=0,1\left(mol\right)\)

\(\Rightarrow\left\{{}\begin{matrix}\%m_{Al}=\dfrac{0,1.27}{40}.100\%=6,75\%\\\%m_{Al_2O_3}=93,25\%\end{matrix}\right.\)

c, \(n_{Al_2O_3}=\dfrac{40.93,25\%}{102}=\dfrac{373}{1020}\left(mol\right)\)

Theo PT: \(n_{HCl}=3n_{Al}+6n_{Al_2O_3}=\dfrac{212}{85}\left(mol\right)\)

\(\Rightarrow V_{HCl}=\dfrac{\dfrac{212}{85}}{2}=\dfrac{106}{85}\left(l\right)\approx1247,06\left(ml\right)\)

d, \(n_{AlCl_3}=n_{Al}+2n_{Al_2O_3}=\dfrac{212}{255}\left(mol\right)\)

\(\Rightarrow m_{AlCl_3}=\dfrac{212}{255}.133,5=\dfrac{9434}{85}\left(g\right)\)

e, \(C_{M_{AlCl_3}}=\dfrac{\dfrac{212}{255}}{\dfrac{106}{85}}=\dfrac{2}{3}\left(M\right)\)

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

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

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

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