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)

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

PTHH :

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

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

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

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

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

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

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

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

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 %

Câu 3 : 

\(n_{Mg}=\dfrac{4,8}{24}=0,2\left(mol\right)\)

a) Pt : \(Mg+2HCl\rightarrow MgCl_2+H_2|\)

            1            2             1          1

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

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

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

\(m_{ddHCl}=\dfrac{14,6.100}{10,95}=133,3\left(g\right)\)

c) \(n_{H2}=\dfrac{0,4.1}{2}=0,2\left(mol\right)\)

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

d) \(n_{MgCl2}=\dfrac{0,2.1}{1}=0,2\left(mol\right)\)

⇒ \(m_{MgCl2}=0,2.95=19\left(g\right)\)

\(m_{ddspu}=4,8+133,3-\left(0,2.2\right)=137,7\left(g\right)\)

\(C_{MgCl2}=\dfrac{19.100}{137,7}=13,8\)⁰/₀

 Chúc bạn học tốt

\(n_{Fe}=\dfrac{2,8}{56}=0,05\left(mol\right)\)

PT :

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

0,05    0,1                       0,05 

\(a,V_{H_2}=0,05.22,4=1,12\left(l\right)\)

\(b,V_{HCl}=\dfrac{n}{C_M}=\dfrac{0,1}{2}=0,05\left(l\right)\)

a)

$n_{Al} = 0,3(mol)$

$2Al + 3H_2SO_4 \to Al_2(SO_4)_3 + 3H_2$

Theo PTHH : 

$n_{H_2SO_4} = \dfrac{3}{2}n_{Al} = 0,45(mol)$
$m_{dd\ H_2SO_4} = \dfrac{0,45.98}{12,25\%} = 360(gam)$

b)

$n_{H_2} = n_{H_2SO_4} = 0,45(mol)$
$V_{H_2} = 0,45.22,4 = 10,08(lít)$

c)

$n_{Al_2(SO_4)_3} = 0,15(mol)$
$m_{dd\ sau\ pư} = 8,1 + 360 - 0,45.2 = 367,2(gam)$
$C\%_{Al_2(SO_4)_3} = \dfrac{0,15.342}{367,2}.100\% = 14\%$

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

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 

\(a)n_{Al}=\dfrac{10,8}{27}=0,4\left(mol\right)\\ 2Al+6HCl\xrightarrow[]{}2AlCl_3+3H_2\\ n_{H_2}=\dfrac{3}{2}n_{Al}=\dfrac{3}{2}\cdot0,4=0,6\left(mol\right)\\ V_{H_2}=0,6.22,4=13,44\left(l\right)\\ b)n_{HCl}=3n_{Al}=3.0,4=1,2\left(mol\right)\\ m_{HCl}=1,2.36,5=43,8\left(g\right)\\ m_{dd_{HCl}}=\dfrac{43,8}{10,95\%}\cdot100\%=400\left(g\right)\\ c)n_{AlCl_3}=n_{Al}=0,4mol\\ m_{AlCl_3}=0,4.133,5=53,4\left(g\right)\\ m_{H_2}=0,6.2=1,2\left(g\right)\\ m_{dd_{AlCl_3}}=10,8+400-1,2=409,6\left(g\right)\\ C_{\%AlCl_3}=\dfrac{53,4}{409,6}\cdot100\%\approx13\%\)