a)V khí H2=3,36

b)m dd=100g

c)C%FeCl2=17,57%

\(n\)Fe = \(\dfrac{8,4}{56}\)= 0,15 mol

Fe + 2HCl -----> FeCl\(2\)+H\(2\)

0,15->0,3 ->0,15 -> 0,15 (mol

V\(H2\) = 0,15 . 22,4 = 3,36 l

b, m_{ct HCl} = 0,3 . 36,5 = 10,95 (g)

m_{dd HCl} = \(\dfrac{10,95}{10,95\%}\) = 100 (g)

c, m_{dd sau pu} = 8,4 + 100 - 0,15.2 = 108,1 g

C_{% FeCl2} = \(\dfrac{0,15.127}{108,1}.100\%\)= 1,76%

n_Fe = \(\dfrac{8,4}{56}\) = 0,15 mol

Fe + 2HCl -> FeCl₂ + H₂

1mol 2mol 1mol 1mol

0,15mol 0,3mol 0,15mol 0,15mol

a) V_H2(đktc) = 0,15. 22,4 =3,36 (l)

b) m_HCl = 0,3 . 36,5 = 10,95 (g)

m_ddHCl = \(\dfrac{mct.100\text{%}}{C\%}\) = \(\dfrac{10,95.100\%}{10,95\%}\)

= 100(g)

c) m_FeCl2 = 127 . 0,15 = 19,05 (g)

m_H2 = 0,15 . 2 = 0,3 (g)

m_dd = (100 + 8,4) - 0,3 =108,1 (g)

C% _FeCl2 = \(\dfrac{mct}{mdd}\) . 100% = \(\dfrac{19,05}{108,1}\) . 100%

= 17,62 %

a, \(Zn+2HCl\rightarrow ZnCl_2+H_2\)

b, \(n_{Zn}=\dfrac{13}{65}=0,2\left(mol\right)\)

Theo PT: \(n_{H_2}=n_{Zn}=0,2\left(mol\right)\Rightarrow V_{H_2}=0,2.22,4=4,48\left(l\right)\)

c, \(n_{HCl}=2n_{Zn}=0,4\left(mol\right)\Rightarrow m_{HCl}=0,4.36,5=14,6\left(g\right)\)

\(\Rightarrow m_{ddHCl}=\dfrac{14,6}{10,95\%}=\dfrac{400}{3}\left(g\right)\)

d, \(n_{ZnCl_2}=n_{Zn}=0,2\left(mol\right)\)

Ta có: m _{dd sau pư} = 13 + 400/3 - 0,2.2 = 2189/15 (g)

\(\Rightarrow C\%_{ZnCl_2}=\dfrac{0,2.136}{\dfrac{2189}{15}}.100\%\approx18,64\%\)

a. 

PTHH:

 Zn + 2HCl ---> ZnCl2 + H2

0.2      0.4           0.2       0.2 (mol)

b.

 nZn=13/65=0.2(mol)

 V H2 = 0.2*22.4 = 4.48 (l)

c.

 mHCl=0.4*36.5=14.6(g)

 mddHCl=14.6/10.95*100~133(g)

d.

 mZn=0.2*35.5=7.1(g)

 mZnCl2=0.2*106=21.2(g)

 mH2=0.2*2=0.4(g)

 Theo ĐLBTKL, ta có: 

   mZn + mddHCl = mddZnCl2 + mH2

   7.1 + 133 = mddZnCl2 + 4

 => mddZnCl2= 7.1 + 133 - 4 = 136.1 (g)

   S ZnCl2= 21.2/136.1*100 ~ 15 (g)

\(n_{Fe}=\dfrac{8,4}{56}=0,15\left(mol\right)\)

PTHH :

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

0,15   0,3          0,15       0,15 

\(V_{H_2}=n.22,4=0,15.22,4=3,36\left(l\right)\)

Còn lại đề thiếu dữ kiện , bạn bổ sung và nếu cần thì đăng lại nha 

\(n_{Zn}=\dfrac{13}{65}=0,2\left(mol\right)\)

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

         1         2            1           1

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

a) \(n_{H2}=\dfrac{0,2.1}{1}=0,2\left(mol\right)\)

\(V_{H2\left(dktc\right)}=0,2.22,4=4,48\left(l\right)\)

b) \(n_{ZnCl2}=\dfrac{0,2.1}{1}=0,2\left(mol\right)\)

⇒ \(m_{ZnCl2}=0,2.136=27,2\left(g\right)\)

c) \(n_{HCl}=\dfrac{0,2.2}{1}=0,4\left(mol\right)\)

⇒ \(m_{HCl}=0,4.36,5=14,6\left(g\right)\)

\(C_{ddHCl}=\dfrac{14,6.100}{250}=5,84\)⁰/₀

 Chúc bạn học tốt

PTHH: \(K_2SO_3+2HCl\rightarrow2KCl+H_2O+SO_2\uparrow\)

a+b+c) Ta có: \(n_{SO_2}=\dfrac{5,6}{22,4}=0,25\left(mol\right)\)

\(\Rightarrow\left\{{}\begin{matrix}n_{KCl}=0,5\left(mol\right)=n_{HCl}\\n_{K_2SO_3}=0,25\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}m_{K_2SO_3}=0,25\cdot158=39,5\left(g\right)=a\\m_{KCl}=0,5\cdot74,5=37,25\left(g\right)\\m_{ddHCl}=\dfrac{0,5\cdot36,5}{10,95\%}\approx166,67\left(g\right)\end{matrix}\right.\)

Mặt khác: \(m_{SO_2}=0,25\cdot64=16\left(g\right)\)

\(\Rightarrow m_{dd}=m_{K_2SO_3}+m_{ddHCl}-m_{SO_2}=190,17\left(g\right)\) \(\Rightarrow C\%_{KCl}=\dfrac{37,25}{190,17}\cdot100\%\approx19,59\%\)

d) PTHH: \(NaOH+HCl\rightarrow NaCl+H_2O\)

Theo PTHH: \(n_{NaOH}=n_{HCl}=0,5\left(mol\right)\) \(\Rightarrow V_{NaOH}=\dfrac{0,5}{0,5}=1\left(l\right)\)

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%

\(n_{Zn}=\dfrac{26}{65}=0,4\left(mol\right)\\ pthh:Zn+2HCl\rightarrow ZnCl_2+H_2\uparrow\) 
           0,4       0,8        0,4           0,4
\(a,V_{H_2}=0,4.22,4=8,96\left(l\right)\\ b,C\%_{HCl}=\dfrac{0,8.36,5}{150}.100\%=19,5\%\\ c,m_{\text{dd}}=26+150-\left(0,4.2\right)=175,2\left(g\right)\\ C\%_{ZnCl_2}=\dfrac{0,4.136}{175,2}.100\%=31\%\)

giup mk voi

\(n_{Al}=\dfrac{10,8}{27}=0,4\left(mol\right)\)

PTHH :

\(2Al+6HCl\rightarrow2AlCl_3+3H_2\uparrow\)

0,4      1,2           0,4         0,6 

\(a,V_{H_2}=0,6.22,4=13,44\left(l\right)\)

\(b,m_{HCl}=1,2.36,5=43,8\left(g\right)\)

\(c,m_{AlCl_3}=133,5.0,4=53,4\left(g\right)\)

\(m_{ddHCl}=\dfrac{43,8.100}{10}=438\left(g\right)\)

\(m_{ddAlCl_3}=10,8+438-\left(0,6.2\right)=447,6\left(g\right)\)

\(C\%=\dfrac{53,8}{447,6}.100\%\approx12,02\%\)