a) 2Al + 6HCl --> 2AlCl₃ + 3H₂

b) \(n_{Al}=\dfrac{8,1}{27}=0,3\left(mol\right)\)

PTHH: 2Al + 6HCl --> 2AlCl₃ + 3H₂

_____0,3--->0,9---------------->0,45

=> V_H2 = 0,45.22,4 = 10,08(l)

c) 

\(C\%\left(HCl\right)=\dfrac{0,9.36,5}{100}.100\%=32,85\%\)

nAl=\(\dfrac{5,4}{27}\)=0,2 mol

2Al+3H₂SO₄→Al₂(SO₄)₃+3H₂

0,2-----0,3---------0,1-----------0,3

=>VH2=0,3.22,4=6,72l

=>CMH2SO4=\(\dfrac{0,3}{0,1}\)=3M

=>CM Al2(SO4)3=\(\dfrac{0,1}{0,1}\)=1M

nAl=\(\dfrac{5,4}{27}\)=0,2 mol

2Al+3H₂SO₄→Al₂(SO₄)₃+3H₂

0,2-----0,3---------0,1-----------0,3

=>VH2=0,3.22,4=6,72l

=>CMH2SO4=\(\dfrac{0,3}{0,1}\)=3M

=>CM Al2(SO4)3=\(\dfrac{0,1}{0,1}\)=1M

Cho e hỏi CM là j ạ

\(m_{H_2}=\dfrac{7,437}{24,79}=0,3mol\\ 2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\)

0,2        0,3               0,1                0,3

\(m_{Al}=0,2.27=5,4g\\ V_{ddH_2SO_4}=\dfrac{0,3}{0,5}=0,6l\\ C_{M_{Al_2\left(SO_4\right)_3}}=\dfrac{0,1}{0,6}=\dfrac{1}{6}\approx0,167M\)

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

b) \(n_{Al}=\dfrac{5,4}{27}=0,2mol\)

\(Theopt:n_{H_2}=\dfrac{3}{2}n_{Al}=0,3mol\)

\(\Rightarrow V_{H_2}=0,3.22,4=6,72lít\)

c) \(Theopt:n_{HCl}=3n_{Al}=0,6mol\)

\(\Rightarrow C_Mdd_{HCl}=\dfrac{0,6}{0,2}=3M\)

\(n_{Al}=\dfrac{5,4}{27}=0,2mol\)

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

0,2                            0,1                0,3  ( mol )

\(m_{Al_2\left(SO_4\right)_3}=0,1.342=34,2g\)

\(V_{H_2}=0,3.22,4=6,72l\)

\(n_{BaCl_2}=\dfrac{200.20,8\%}{208}=0,2\left(mol\right)\\ PTHH:BaCl_2+H_2SO_4\rightarrow BaSO_4+2HCl\\ n_{BaSO_4}=n_{H_2SO_4}=n_{BaCl_2}=0,2\left(mol\right)\\ a,m_{kt}=m_{BaSO_4}=233.0,2=46,6\left(g\right)\\ b,C\%_{ddH_2SO_4}=\dfrac{0,2.98}{200}.100\%=9,8\%\)

a) \(n_{Al}=\dfrac{8,1}{27}=0,3\left(mol\right)\)

PTHH: 2Al + 6HCl --> 2AlCl₃ + 3H₂

           0,3-->0,9------>0,3--->0,45

=> \(V_{dd.HCl}=\dfrac{0,9}{1,5}=0,6\left(l\right)\)

b) \(C_{M\left(AlCl_3\right)}=\dfrac{0,3}{0,6}=0,5M\)

\(n_{K_2SO_3}=\dfrac{15.8}{158}=0.1\left(mol\right)\)

\(K_2SO_3+H_2SO_4\rightarrow K_2SO_4+SO_2+H_2O\)

\(0.1...........................0.1..........0.1\)

\(V_{SO_2}=0.1\cdot22.4=2.24\left(l\right)\)

\(C_{M_{K_2SO_4}}=\dfrac{0.1}{0.2}=0.5\left(M\right)\)

a)

$K_2SO_3 + H_2SO_4 \to K_2SO_4 + SO_2 + H_2O$
b)

$n_{SO_2} = n_{H_2SO_4} = n_{K_2SO_3} = \dfrac{15,8}{158} = 0,1(mol)$
$V_{SO_2} = 0,1.22,4 = 2,24(lít)$
c)

$C_{M_{H_2SO_4}} = \dfrac{0,1}{0,2} = 0,5M$