Zn+2HCl->ZnCl2+H2

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

nZn=0,2 mol

=>m Hcl=0,4.36,5=14,6g

m muối=0,2.136=27,2g

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

`Zn + 2HCl -> ZnCl_2 + H_2` `\uparrow`

`n_(Zn) = 13/65 = 0,2 mol`.

`n_(HCl) = 0,4 mol`.

`m_(HCl) = 0,4 xx 36,5 = 14,6g`.

c, `m_(ZnCl_2) = 0,2 xx 127 = 25,4 g`.

`d, V_(H_2) = 0,2 xx 22,4 = 4,48l`.

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

 Theo PTHH : n H2 = n Fe = 8,4/56 = 0,15(mol)

V H2 = 0,15.22,4 = 3,36(lít)

b) n HCl = 2n Fe = 0,3(mol)

=> CM HCl = 0,3/0,2 = 1,5M

c) $CuO + H_2 \xrightarrow{t^o} Cu + H_2O$

Ta thấy :

n CuO = 32/80 = 0,4 > n H2 = 0,15 mol nên CuO dư

Theo PTHH : n Cu = n H2 = 0,15 mol

=> m Cu = 0,15.64 = 9,6 gam

Ta có: \(n_{Fe}=\dfrac{5,6}{56}=0,1\left(mol\right)\)

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

___0,1______________0,1____0,1 (mol)

\(FeSO_4+BaCl_2\rightarrow FeCl_2+BaSO_{4\downarrow}\)

__0,1______0,1_____________0,1 (mol)

a, V_H2 = 0,1.22,4 = 2,24 (l)

b, \(V_{BaCl_2}=\dfrac{0,1}{0,2}=0,5\left(l\right)\)

c, \(m_{BaSO_4}=0,1.233=23,3\left(g\right)\)

Bạn tham khảo nhé!

PTHH: \(Zn+2HCl\rightarrow ZnCl_2+H_2\uparrow\)

Ta có: \(n_{Zn}=\dfrac{32,5}{65}=0,5\left(mol\right)\)

\(\Rightarrow\left\{{}\begin{matrix}n_{HCl}=1\left(mol\right)\\n_{H_2}=0,5\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}m_{ddHCl}=\dfrac{1\cdot36,5}{20\%}=182,5\left(g\right)\\V_{H_2}=0,5\cdot22,4=11,2\left(l\right)\end{matrix}\right.\)

a.\(n_{H_2}=\dfrac{V_{H_2}}{22,4}=\dfrac{8,96}{22,4}=0,4mol\)

Gọi \(\left\{{}\begin{matrix}n_{Fe}=x\\n_{Zn}=y\end{matrix}\right.\)

\(Fe+2HCl\rightarrow FeCl_2+H_2\)

 x                                      x ( mol )

\(Zn+2HCl\rightarrow ZnCl_2+H_2\)

 y                                        y   ( mol )

Ta có:

\(\left\{{}\begin{matrix}56x+65y=25,55\\x+y=0,4\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=0,05\\y=0,35\end{matrix}\right.\)

\(\Rightarrow m_{Fe}=0,05.56=2,8g\)

\(\Rightarrow m_{Zn}=0,35.65=22,75g\)

\(\%m_{Fe}=\dfrac{2,8}{25,55}.100=10,95\%\)

\(\%m_{Zn}=100\%-10,95\%=89,05\%\)

b.\(n_{HCl}=2.0,05+2.0,35=0,8mol\)

\(C_M=\dfrac{n}{V}\Rightarrow V=\dfrac{n}{C_M}=\dfrac{0,8}{2}=0,4l\)

a, \(CO_2+Ca\left(OH\right)_2\rightarrow CaCO_3+H_2O\)

b, \(n_{CO_2}=\dfrac{5,6}{22,4}=0,25\left(mol\right)\)

\(n_{Ca\left(OH\right)_2}=n_{CO_2}=0,25\left(mol\right)\Rightarrow C_{M_{Ca\left(OH\right)_2}}=\dfrac{0,25}{0,2}=1,25\left(M\right)\)

c, \(Ca\left(OH\right)_2+2HCl\rightarrow CaCl_2+2H_2O\)

Theo PT: \(n_{HCl}=2n_{Ca\left(OH\right)_2}=0,5\left(mol\right)\)

\(\Rightarrow m_{ddHCl}=\dfrac{0,5.36,5}{15\%}\approx121,67\left(g\right)\)

\(n_{CO_2}=\dfrac{5,6}{22,4}=0,25mol\\ a)CO_2+Ca\left(OH\right)_2\rightarrow Ca\left(OH\right)_2+H_2O\)

    0,25       0,25                0,25

\(b)C_{M_{Ca\left(OH\right)_2}}=\dfrac{0,25}{0,2}=1,25M\\ c)2HCl+Ca\left(OH\right)_2\rightarrow CaCl_2+2H_2O\\ n_{HCl}=2n_{Ca\left(OH\right)_2}=2.0,25=0,5mol\\ m_{ddHCl}=\dfrac{0,5.36,5}{15\%}\cdot100\%\approx121,67g\)

\(n_{HCl}=0,3.2=0,6\left(mol\right)\\a, 2Al+6HCl\rightarrow2AlCl_3+3H_2\\ b,n_{H_2}=\dfrac{3}{6}.0,6=0,3\left(mol\right)\\ V_{H_2\left(đktc\right)}=0,3.22,4=6,72\left(l\right)\\ c,n_{Al}=n_{AlCl_3}=\dfrac{2}{6}.0,6=0,2\left(mol\right)\\ m_{Al}=0,2.27=5,4\left(g\right)\\ d,V_{ddAlCl_3}=V_{ddHCl}=0,3\left(l\right)\\ C_{MddHCl}=\dfrac{0,2}{0,3}=\dfrac{2}{3}\left(M\right)\)

nZn = 13/65 = 0.2 (mol) 

Zn + H2SO4 => ZnSO4 + H2

0.2......0.2..........................0.2 

VH2 = 0.2*22.4 = 4.48 (l)

C%H2SO4 = 0.2*98/200 * 100% = 9.8 %

nCuO = 8/80 = 0.1 (mol) 

CuO + H2 -to-> Cu + H2O 

0.1......0.1...........0.1 

=> H2 dư 

mCu = 0.1*64 = 6.4 (g) 

â) nZn=0,2(mol)

PTHH: Zn + H2SO4 -> ZnSO4 + H2

0,2_____0,2______0,2_____0,2(mol)

=> V(H2,đktc)=0,2.22,4=4,48(l)

b) C%ddH2SO4= [(98.0,2)/200)].100=9,8%

c) nCuO=0,1(mol)

PTHH: CuO + H2 -to-> Cu + H2O

Ta có: 0,1/1 < 0,2/1

=> H2 dư, CuO hết, tính theo nCuO

=> nCu=nCuO=0,1(mol)

=>mCu=6,4(g)