n_Mg = 0,1(mol)

PTHH: Mg + 2HCl --> MgCl₂ +H₂

n_Mg = n_MgCl2= n_H2 = 0,1(mol)

=> m_muối = 9,5(g) 

_VH2 = 2,24(l)

b) CM_HCl = 0,2/0,1=2(M)

Bài 8:

\(n_{H_2}=\dfrac{7,437}{24,79}=0,3\left(mol\right)\)

PT: \(2Al+6HCl\rightarrow2AlCl_3+3H_2\)

_____0,2______0,6_____0,2____0,3 (mol)

a, \(m_{Al}=0,2.27=5,4\left(g\right)\)

b, \(C_{M_{HCl}}=\dfrac{0,6}{0,3}=2\left(M\right)\)

c, \(C_{M_{AlCl_3}}=\dfrac{0,2}{0,2}=1\left(M\right)\)

Bài 9:

Ta có: \(n_{H_2}=\dfrac{2,479}{24,79}=0,1\left(mol\right)\)

PT: \(Mg+2HCl\rightarrow MgCl_2+H_2\)

\(MgO+2HCl\rightarrow MgCl_2+H_2O\)

a, \(n_{Mg}=n_{H_2}=0,1\left(mol\right)\Rightarrow m_{Mg}=0,1.24=2,4\left(g\right)\)

\(\Rightarrow m_{MgO}=8,4-2,4=6\left(g\right)\)

b, \(n_{MgO}=\dfrac{6}{40}=0,15\left(mol\right)\)

Theo PT: \(n_{HCl}=2n_{Mg}+2n_{MgO}=0,5\left(mol\right)\)

\(\Rightarrow m_{ddHCl}=\dfrac{0,5.36,5}{3,65\%}==500\left(g\right)\)

\(n_{Zn}=\dfrac{3,25}{65}=0,05\left(mol\right)\)

a) Pt : \(Zn+2HCl\rightarrow ZnCl_2+H_2|\)

            1          2             1           1

         0,05      0,1                       0,05

b) \(n_{H2}=\dfrac{0,05.1}{1}=0,05\left(mol\right)\)

\(V_{H2\left(dktc\right)}=0,05.22,4=1,12\left(l\right)\)

c) \(n_{HCl}=\dfrac{0,05.2}{1}=0,1\left(mol\right)\)

200ml = 0,2l

\(C_{M_{ddHCl}}=\dfrac{0,1}{0,2}=0,5\left(M\right)\)

 Chúc bạn học tốt

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_H2 = 0,2.22,4 = 4,48 (l)

b) m_HCl = 0,4.36,5 = 14,6 (g)

=> \(m_{dd.HCl}=\dfrac{14,6.100}{7,3}=200\left(g\right)\)

c)

m_{dd sau pư} = 13 + 200 - 0,2.2 = 212,6 (g)

m_ZnCl2 = 0,2.136 = 27,2 (g)

=> \(C\%=\dfrac{27,2}{212,6}.100\%=12,8\%\)

\(n_{H_2SO_4}=0,2.2=0,4\left(mol\right)\)

Fe + H2SO4 → FeSO4 + H2 

2Al + 3H2SO4 → Al2(SO4)3 + 3H2

Gọi x,y lần lượt là số mol Fe, Al

\(\left\{{}\begin{matrix}56x+27y=11\\x+\dfrac{3}{2}y=0,4\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x=0,1\\y=0,2\end{matrix}\right.\)

=>\(\%m_{Fe}=\dfrac{0,1.56}{11}.100=50,91\%\)

=> %m _Al = 100 - 50,91 =49,09 %

b)Theo PT:  \(n_{H_2}=n_{H_2SO_4}=0,4\left(mol\right)\)

=> \(V_{H_2}=0,4.22,4=8,96\left(l\right)\)

c) \(CM_{FeSO_4}=\dfrac{0,1}{0,2}=0,5M\)

\(CM_{Al_2\left(SO_4\right)_3}=\dfrac{\dfrac{0,2}{2}}{0,2}=0,5M\)

nH2=13,14:22,4=0,6 mol

PTHH: 2Al+6HCl=>2Al2Cl3+3H2

             0,4<-1,2<----0,4<-----0,6

=> Al=0,4.27=10,8g

CMHCL=1,2:0,4=3M

CM Al2Cl3=0,4:0,4=1M

bài 2: nH2=0,2mol

PTHH: 2A+xH2SO4=> A2(SO4)x+xH2

             0,4:x<---------------------------0,2

ta có PT: \(\frac{13}{A}=\frac{0,4}{x}\)<=> 13x=0,4A

=> A=32,5x

ta lập bảng xét

x=1=> A=32,5   loiaj

x=2=> A=65    nhận

x=3=> A=97,5    loại

=> A là kẽm (Zn) 

a) $2Al + 6HCl \to 2AlCl_3 + 3H_2$

b) n Al = 8,1/27 = 0,3(mol)

Theo PTHH : 

n H2 = 3/2 n Al = 0,45(mol)

V H2 = 0,45.22,4 = 10,08(lít)

c) n AlCl3 = n Al = 0,3(mol)

m AlCl3 = 0,3.133,5 = 40,05(gam)

d) n HCl = 3n Al = 0,9(mol)

m dd HCl = 0,9.36,5/7,3% = 450(gam)

Sau phản ứng : 

m dd = 8,1 + 450  -0,45.2 = 457,2(gam)

C% AlCl3 = 40,05/457,2  .100% = 8,76%

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

              a_______a_______a_____a    (mol)

            \(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\uparrow\)

                2b______3b__________b_____3b    (mol)

Ta lập HPT: \(\left\{{}\begin{matrix}56a+27\cdot2b=11\\a+3b=0,2\cdot2=0,4\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}a=0,1\\b=0,1\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}\%m_{Fe}=\dfrac{0,1\cdot56}{11}\cdot100\%\approx50,91\%\\\%m_{Al}=49,09\%\end{matrix}\right.\)

Theo các PTHH: \(\left\{{}\begin{matrix}n_{H_2}=0,4\left(mol\right)\\n_{FeSO_4}=0,1\left(mol\right)\\n_{Al_2\left(SO_4\right)_3}=0,3\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}V_{H_2}=0,4\cdot22,4=8,96\left(l\right)\\C_{M_{FeSO_4}}=\dfrac{0,1}{0,2}=0,5\left(M\right)\\C_{M_{Al_2\left(SO_4\right)_3}}=\dfrac{0,3}{0,2}=1,5\left(M\right)\end{matrix}\right.\) 

a) nH2SO4=0,4(mol)

Đặt: nFe=x(mol); nAl=y(mol) (x,y>0)

PTHH: Fe + H2SO4 -> FeSO4 + H2

x________x______x______x(mol)

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

y____1,5y_______0,5y_______1,5y(mol)

Ta có hpt:

\(\left\{{}\begin{matrix}56x+27y=11\\x+1,5y=0,4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0,1\\y=0,2\end{matrix}\right.\)

=> mFe=0,1.56=5,6(g)

=>%mFe=(5,6/11).100=50,909%

=>%mAl= 49,091%

b) V(H2,đktc)=0,4.22,4=8,96(l)

c) nAl2(SO4)3= 0,5y=0,5.0,2=0,1(mol)

nFeSO4=x=0,1(mol)

Vddsau=VddH2SO4=0,2(l)

=>CMddAl2(SO4)3= 0,1/0,2=0,5(M)

CMddFeSO4=0,1/0,2=0,5(M)