\(n_{H_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\)

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

\(\left(mol\right)\)       \(0,2\)     \(0,6\)         \(0,2\)           \(0,3\)

\(a.m=0,2.27=5,4\left(g\right)\\ a=\dfrac{36,5.0,6.100}{7,3}=300\left(g\right)\\ b.\)

\(AlCl_3-\) Nhôm clorua

\(m_{AlCl_3}=0,2.133,5=26,7\left(g\right)\\ c.m_{H_2}=0,3.2=0,6\left(g\right)\)

\(n_{H_2}=\dfrac{6.72}{22.4}=0.3\left(mol\right)\)

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

\(0.2........0.6..........0.2.............0.3\)

\(m_{Al}=0.2\cdot27=5.4\left(g\right)\)

\(m_{HCl}=0.6\cdot36.5=21.9\left(g\right)\)

\(\Rightarrow m_{dd_{HCl}}=\dfrac{21.9}{7.3\%}=300\left(g\right)\)

\(m_{AlCl_3}=0.2\cdot133.5=26.7\left(g\right)\)

( Nhôm clorua ) 

\(m_{H_2}=0.3\cdot2=0.6\left(g\right)\)

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)

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

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%

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, \(n_{ZnCl_2}=n_{Zn}=0,2\left(mol\right)\Rightarrow m_{ZnCl_2}=0,2.136=27,2\left(g\right)\)

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

\(\Rightarrow m_{ddHCl}=\dfrac{14,6}{7,3\%}=200\left(g\right)\)

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

\(\Rightarrow C\%_{ZnCl_2}=\dfrac{27,2}{212,6}.100\%\approx12,79\%\)

Bài 2 : 

a)

$Mg + 2HCl \to MgCl_2 + H_2$
$2Al + 6HCl \to 2AlCl_3 + 3H_2$
$n_{H_2} = \dfrac{5,6}{22,4} = 0,25(mol)$
$n_{HCl} = 2n_{H_2} = 0,5(mol)$
$m_{HCl} = 0,5.36,5 = 18,25(gam)$

b)

Bảo toàn khối lượng : 

$m_A = 22,85 + 0,25.2 - 18,25 = 5,1(gam)$

Có: \(n_{Mg}=\dfrac{6}{24}=0,25\left(mol\right)\)

PT: \(Mg+2HCl\rightarrow MgCl_2+H_2\)

___0,25___________0,25____0,25 (mol)

a, \(V_{H_2}=0,25.22,4=5,6\left(l\right)\)

b, \(m_{MgCl_2}=0,25.95=23,75\left(g\right)\)

Bạn tham khảo nhé!

Đặt : 

nAl = a (mol) 

nFe = b(mol) 

mX = 27a + 56b = 16.6 (g) (1) 

2Al + 3H2SO4 => Al2(SO4)3 + 3H2 

Fe + H2SO4 => FeSO4 + H2 

mM = 342a  + 152b = 64.6 (g) (2) 

(1) , (2):

a = 4/55

b = 23/88

%Al = (4/55*27) / 16.6 *100% = 11.83%

%Fe = 100 - 11.83 = 88.17%

nH2 = 3/2a + b = 3/2 * 4/55 + 23/88 = 163/440 (mol) 

VH2 = 8.3 (l)