Tóm tắt: m_Al=5,4(g)

Tính m_HCl ; m_AlCl3 ; V_H2

Giải: n_Al=\(\dfrac{m}{M}=\dfrac{5,4}{27}=0,2\left(mol\right)\)

a, pthh: 2Al + 6HCl \(\rightarrow\) 2AlCl₃ + 3H₂

0,2........0,6..........0,2...........0,3 (mol)

b, m_HCl=n.M=0,6.36,5=21,9(g)

m_AlCl3=n.M=0,2.133,5=26,7(g)

V_H2=n.22,4=0,3.22,4=6,72(lít)

1.

n_Fe = 0,1 mol

Fe + 2HCl \(\rightarrow\) FeCl₂ + H₂

\(\Rightarrow\) V_H2 = 0,1.22,4 = 2,24 (l)

\(\Rightarrow\) m_FeCl2 = 0,1.127 = 12,7 (g)

n_Al = 0,2 mol

n_HCl = 0,3 mol

2Al + 6HCl \(\rightarrow\) 2AlCl₃ + 3H₂

Đặt tỉ lệ ta có

\(\dfrac{0,2}{2}\) > \(\dfrac{0,3}{6}\)

\(\Rightarrow\) Al dư

\(\Rightarrow\) V_H2 = 0,15.22,4 = 3,36 (l)

\(\Rightarrow\) m_AlCl3 = 0,1.133,5 = 13,35 (g)

1.

a) PTHH :

Fe + 2HCl FeCl2 + H2

b) theo pt : mFe = 5,6 g

=> nFe = 0,1 mol

pthh:

Fe + 2HCl FeCl2 + H2

1mol..2mol.....1mol....1mol

0,1mol.0,2mol..0,1mol..0,1mol

V_H2 = 0,1.22,4 = 2,24 (l)

b) ta có n_FeCl2 = 0,1mol

m_FeCl2 = 0,1.127 = 12,7 (g)

\(n_{HCl}=\dfrac{14,6}{36,5}=0,4\left(mol\right)\)

\(PTHH:R+2HCl\rightarrow RCl_2+H_2\uparrow\\ Mol:0,2\leftarrow0,4\rightarrow0,2\rightarrow0,2\)

=> M_R = \(\dfrac{13}{0,2}=65\left(\dfrac{g}{mol}\right)\)

=> R là Zn

=> \(\left\{{}\begin{matrix}V_{H_2}=0,2.22,4=4,48\left(l\right)\\m_{ZnCl_2}=0,2.136=27,2\left(g\right)\end{matrix}\right.\)

nHCL = 14,6  : 36,5 = 0,4 (MOL) 
pthh :  2R + 2xHCl ---> 2RClx  + xH2 
            0,4x<--0,4 (mol) 
M_R =  13:0,4x = 32,5x(g/mol) 
xét 
x = 1 (KTM ) 
x= 2 (TM ) 
x = 3 (KTM ) 
x =4( KTM ) 
x= 5 (ktm ) 
x=6 (ktm) 
x=7 (ktm ) 
=> R là zn

nHCL = 14,6  : 36,5 = 0,4 (MOL) 
pthh :  2R + 2xHCl ---> 2RClx  + xH2 
            0,4x<--0,4 (mol) 
MR =  13:0,4x = 32,5x(g/mol) 
xét 
x = 1 (KTM ) 
x= 2 (TM ) 
x = 3 (KTM ) 
x =4( KTM ) 
x= 5 (ktm ) 
x=6 (ktm) 
x=7 (ktm ) 
=> R là zn

a)

\(Mg + 2HCl \to MgCl_2 + H_2\\ 2Al + 6HCl \to 2AlCl_3 + 3H_2\\ Fe + 2HCl \to FeCl_2 + H_2\\ Zn + 2HCl \to ZnCl_2 + H_2\)

b)

Gọi : \(n_{H_2} = a(mol) \Rightarrow n_{HCl} = 2a\)

Bảo toàn khối lượng :

\(13,5 + 2a.36,5 = 66,75 + 2.a\\ \Rightarrow a = 0,75\\ \Rightarrow V = 0,75.22,4 = 16,8(lít)\)

a) Mg + 2 HCl -> MgCl2  + H2

2Al + 6 HCl -> 2 AlCl3 + 3 H2

Fe + 2 HCl ->  FeCl2 + H2

Zn + 2 HCl -> ZnCl2 + H2

b) mY-mX=mCl

<=> mCl= 66,75-13,5=53,25(g)

=>nCl=53,25/35,5=1,5(mol)

=> nH2= nCl/2= 1,5/2=0,75(mol)

=>V=V(H2,đktc)=0,75.22,4=16,8(l)

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

PTHH: Fe + 2HCl --> FeCl₂ + H₂

          0,2------------>0,2----->0,2

=> \(\left\{{}\begin{matrix}m_{FeCl_2}=0,2.127=25,4\left(g\right)\\V_{H_2}=0,2.22,4=4,48\left(l\right)\end{matrix}\right.\)

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

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: \(n_{Fe}=\dfrac{11.2}{56}=0.2\left(mol\right)\)

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

0,2      0,4        0,2         0,2

\(V_{H_2}=0,2\cdot22,4=4,48\left(lít\right)\)

b: \(\dfrac{n_{HCl}}{V_{HCl}}=2\)

=>\(\dfrac{0.4}{V_{HCl}}=2\)

=>\(V_{HCl}=\dfrac{0.4}{2}=0.2\left(lít\right)\)

c: \(C_M=\dfrac{n}{V}=\dfrac{0.2}{0.2}=1\)