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

Pt : \(Fe+H_2SO_4\rightarrow FeSO_4+H_2|\)

        1           1              1          1

      0,1       0,1            0,1        0,1

a) \(n_{H2}=\dfrac{0,1.1}{1}=0,1\left(mol\right)\)

\(V_{H2\left(dktc\right)}=0,1.22,4=2,24\left(l\right)\)

b) \(n_{FeSO4}=\dfrac{0,1.1}{1}=0,1\left(mol\right)\)

⇒ \(m_{FeSO4}=0,1.152=15,2\left(g\right)\)

c) \(n_{H2SO4}=\dfrac{0,1.1}{1}=0,1\left(mol\right)\)

\(m_{H2SO4}=0,1.98=9,8\left(g\right)\)

\(m_{ddH2SO4}=\dfrac{9,8.100}{10}=98\left(g\right)\)

 Chúc bạn học tốt

\(n_{Al}=\dfrac{2,7}{27}=0,1\left(mol\right)\)

Pt : \(2Al++3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2|\)

         2                3                 1               3

       0,1              0,15           0,05           0,15

a) \(n_{H2}=\dfrac{0,1.3}{2}=0,15\left(mol\right)\)

\(V_{H2\left(dktc\right)}=0,15.22,4=3,36\left(l\right)\)

b) \(n_{Al2\left(SO4\right)3}=\dfrac{0,15.1}{3}=0,05\left(mol\right)\)

⇒ \(m_{Al2\left(SO4\right)3}=0,05.342=17,1\left(g\right)\)

c) \(n_{H2SO4}=\dfrac{0,1.3}{2}=0,15\left(mol\right)\)

200ml = 0,2l

\(C_{M_{ddH2SO4}}=\dfrac{0,15}{0,2}=0,75\left(M\right)\)

 Chúc bạn học tốt

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

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)

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

\(n_{Al}=\dfrac{6,75}{27}=0,25\left(mol\right)\)

PTHH :

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

0,25    0,75      0,25          0,375 

\(a,V_{H_2}=0,375.22,4=8,4\left(l\right)\)

\(b,m_{HCl}=0,75.36,5=27,375\left(g\right)\)

\(m_{ddHCl}=\dfrac{27,375.100}{10,95}=250\left(g\right)\)

\(c,m_{AlCl_3}=0,25.133,5=33,375\left(g\right)\)

\(m_{ddAlCl_3}=6,75+250-\left(0,375.2\right)=256\left(g\right)\)

\(C\%_{AlCl_3}=\dfrac{33,375}{256}.100\%\approx13,04\left(\%\right)\)

\(n_{Fe}=\dfrac{84}{56}=1,5\left(mol\right)\\ PTHH:Fe+2HCl\rightarrow FeCl_2+H_2\\ n_{H_2}=n_{FeCl_2}=n_{Fe}=1,5\left(mol\right)\\ V_{H_2}=1,5.22,4=33,6\left(l\right)\\ C\%_{ddFeCl_2}=\dfrac{127.1,5}{84+300-1,5.2}.100\%=\dfrac{190,5}{381}.100\%=50\%\)

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)

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 %