nAl = 5.4/27 = 0.2 mol 

2Al + 3H2SO4 => Al2(SO4)3 + 3H2 

0.2______________0.1________0.3 

VH2 = 0.3*22.4 = 6.72 (l) 

mAl2(SO4)3 = 0.1*342 = 34.2 g 

\(n_{Al}=\dfrac{1,35}{27}=0,05\left(mol\right)\\ 2Al+3H_2SO_4\rightarrow2Al_2\left(SO_4\right)_3+3H_2\\ n_{H_2SO_4}=n_{H_2}=\dfrac{3}{2}.0,05=0,075\left(mol\right)\\ n_{Al_2\left(SO_4\right)_3}=\dfrac{0,05}{2}=0,025\left(mol\right)\\ a,m_{Al_2\left(SO_4\right)_3}=342.0,025=8,55\left(g\right)\\ b,V_{H_2\left(đktc\right)}=0,075.22,4=1,68\left(l\right)\\ c,m_{H_2SO_4}=0,075.98=7,35\left(g\right)\)

\(n_{Al}=\dfrac{1,35}{27}=0,05mol\)

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

0,05      0,075        0,025           0,075

\(m_{Al_2\left(SO_4\right)_3}=0,025\cdot342=8,55g\)

\(V_{H_2}=0,075\cdot22,4=1,68l\)

\(m_{H_2SO_4}=0,075\cdot98=7,35g\)

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

b)

$n_{H_2SO_4} = \dfrac{200.15\%}{98} = \dfrac{15}{49}(mol)$

Theo PTHH :

$n_{H_2} = n_{H_2SO_4} = \dfrac{15}{49}(mol)$
$n_{Al_2(SO_4)_3} = \dfrac{1}{3}n_{H_2SO_4} = \dfrac{5}{49}(mol)$

Vậy :

$V_{H_2} = \dfrac{15}{49}.22,4 = 6,86(lít)$
$m_{Al_2(SO_4)_3} = \dfrac{5}{49}.342 = 34,9(gam)$

\(n_{H_2SO_4}=\dfrac{200\cdot15\%}{98}=\dfrac{15}{49}\left(mol\right)\)

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

\(........\dfrac{15}{49}.........\dfrac{5}{49}......\dfrac{15}{49}\)

\(m_{Al_2\left(SO_4\right)_3}=\dfrac{5}{49}\cdot342=35\left(g\right)\)

\(V_{H_2}=\dfrac{15}{49}\cdot22.4=6.85\left(l\right)\)

a, \(PTHH:3H_2SO_4+2Al\rightarrow Al_2\left(SO_4\right)_3+3H_2\)

b, Ta có : \(n_{H2SO4}=\dfrac{m}{M}=\dfrac{15}{49}\left(mol\right)\)

Theo PTHH : \(n_{Al2\left(SO4\right)3}=\dfrac{1}{3}n_{H2SO4}=\dfrac{5}{49}\left(mol\right)\)

\(\Rightarrow m_{Al2\left(SO4\right)3}=n.M\approx34,9g\)

Theo PTHH : \(n_{H2}=n_{H2SO4}=\dfrac{15}{49}\left(mol\right)\)

\(\Rightarrow V=6,85\left(l\right)\)

Đề sai nha bạn

a.b.\(n_{Al}=\dfrac{m_{Al}}{M_{Al}}=\dfrac{5,4}{27}=0,2mol\)

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

0,2                                              0,3    ( mol )

\(V_{H_2}=n_{H_2}.22,4=0,3.22,4=6,72l\)

c.\(n_{CuO}=\dfrac{m_{CuO}}{M_{CuO}}=\dfrac{32}{80}=0,4mol\)

\(CuO+H_2\rightarrow Cu+H_2O\)

 0,4  <  0,3                          ( mol )

0,3        0,3      0,3                   ( mol )

\(m_A=m_{CuO\left(du\right)}+m_{Cu}=\left[\left(0,4-0,3\right).80\right]+\left(0,3.64\right)=8+19,2=27,2g\)

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

Làm gộp các phần còn lại

Ta có: \(n_{Al}=\dfrac{5,4}{27}=0,2\left(mol\right)\) \(\Rightarrow\left\{{}\begin{matrix}n_{Al_2\left(SO_4\right)_3}=0,1mol\\n_{H_2SO_4}=n_{H_2}=0,3mol\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}V_{H_2}=0,3\cdot22,4=6,72\left(l\right)\\m_{Al_2\left(SO_4\right)_3}=0,1\cdot342=34,2\left(g\right)\\m_{H_2SO_4}=0,3\cdot98=29,4\left(g\right)\end{matrix}\right.\)