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

b: \(n_{Al}=\dfrac{2.5}{27}< \dfrac{1}{4}\)

=>H₂SO₄ dư, Al đủ

\(m_{H_2SO_4}=0.25\cdot98=24.5\left(g\right)\)

c: \(n_{Al_2\left(SO_4\right)_3}=\dfrac{2.5}{54}=\dfrac{5}{108}\left(mol\right)\)

\(\Leftrightarrow n_{H_2}=\dfrac{5}{36}\left(mol\right)\)

\(V_{H_2}=\dfrac{5}{36}\cdot22.4=\dfrac{28}{9}\left(lít\right)\)

Mình thấy bạn Thịnh tính lượng dư sai

Đây là bài mình từng làm, bạn tham khảo nhé!

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

Làm gộp các phần còn lại

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

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

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

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

Xét tỉ lệ: \(\dfrac{0,1}{2}>\dfrac{0,25}{6}\), ta được Al dư.

Theo PT: \(n_{H_2}=\dfrac{1}{2}n_{HCl}=0,125\left(mol\right)\Rightarrow V_{H_2}=0,125.22,4=2,8\left(l\right)\)

b, Theo PT: \(n_{Al\left(pư\right)}=\dfrac{1}{3}n_{HCl}=\dfrac{1}{12}\left(mol\right)\Rightarrow n_{Al\left(dư\right)}=0,1-\dfrac{1}{12}=\dfrac{1}{60}\left(mol\right)\)

\(\Rightarrow m_{Al}=\dfrac{1}{60}.27=0,45\left(g\right)\)

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

\(n_{H_2SO_4}=\dfrac{39,2}{98}=0,4mol\)

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

Xét: \(\dfrac{0,2}{2}\) < \(\dfrac{0,4}{3}\)                                    ( mol )

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

\(m_{H_2SO_4\left(dư\right)}=\left(0,4-0,3\right).98=9,8g\)

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

c.\(2H_2+O_2\rightarrow\left(t^o\right)2H_2O\)   

   0,3     0,15                        ( mol )

\(V_{kk}=V_{O_2}.5=\left(0,15.22,4\right).5=16,8l\)

Chúc bạn học tốt!

\(a) 2Al + 6HCl \to 2AlCl_3 + 3H_2\\ b)n_{AlCl_3} = n_{Al} = \dfrac{8,1}{27} = 0,3(mol)\\ \Rightarrow m_{AlCl_3} = 0,3.133,5 = 40,05(gam)\\ c) n_{HCl} = 3n_{Al} = 0,9(mol)\\ \Rightarrow C_{M_{HCl}} = \dfrac{0,9}{0,5} = 1,8M\)

\(n_{Fe}=\dfrac{11,2}{56}=0,2\left(mol\right)\)

PTHH : Fe + 2HCl -> FeCl₂ + H₂

           0,2      0,4                      0,2

Xét tỉ lệ \(\dfrac{0,2}{1}< \dfrac{0,5}{2}\) => Fe đủ , HCl dư

\(m_{HCl\left(dư\right)}=\left(0,5-0,4\right).36,5=3,65\left(g\right)\)

\(V_{H_2}=0,2.22,4=4,48\left(l\right)\)

11,2 g j bạn 

a) $Zn + H_2SO_4 \to ZnSO_4 + H_2$

b) Theo PTHH :

$n_{H_2} = n_{ZnSO_4} = n_{Zn} = \dfrac{32,5}{65} = 0,5(mol)$

$V_{H_2} = 0,5.22,4 = 11,2(lít)$

$m_{ZnSO_4} = 0,5.161 = 80,5(gam)$

c) 

$n_{H_2SO_4} = n_{Zn} = 0,5(mol)$

$C\%_{H_2SO_4} = \dfrac{0,5.98}{168,5}.100\% = 29,08\%$