a, PT: \(Fe+H_2SO_4\rightarrow FeSO_4+H_2\)

b, Ta có: \(n_{Fe}=\dfrac{19,6}{56}=0,35\left(mol\right)\)

Theo PT: \(n_{H_2}=n_{Fe}=0,35\left(mol\right)\)

\(\Rightarrow V_{H_2}=0,35.22,4=7,84\left(l\right)\)

c, Ta có: \(n_{CuO}=\dfrac{12}{80}=0,15\left(mol\right)\)

PT: \(CuO+H_2\underrightarrow{t^o}Cu+H_2O\)

Xét tỉ lệ: \(\dfrac{0,15}{1}< \dfrac{0,35}{1}\), ta được H₂ dư.

Theo PT: \(n_{H_2\left(pư\right)}=n_{CuO}=0,15\left(mol\right)\)

\(\Rightarrow n_{H_2\left(dư\right)}=0,35-0,15=0,2\left(mol\right)\)

\(\Rightarrow m_{H_2\left(dư\right)}=0,2.2=0,4\left(g\right)\)

\(n_{Al}=\dfrac{2,5}{27}=\dfrac{25}{270}=\dfrac{5}{54}\left(mol\right)\\ n_{H_2SO_4}=0,5\left(mol\right)\\ a,2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\\ b,Vì:\dfrac{\dfrac{5}{54}}{2}< \dfrac{0,5}{4}\Rightarrow H_2SO_4dư\\ b,n_{H_2SO_4\left(dư\right)}=0,5-\dfrac{3}{2}.\dfrac{5}{54}=\dfrac{13}{36}\left(mol\right)\\ \Rightarrow m_{H_2SO_4}=\dfrac{13}{36}.98=\dfrac{637}{18}\left(g\right)\\ c,n_{H_2}=\dfrac{3}{2}.n_{Al}=\dfrac{3}{2}.\dfrac{5}{54}=\dfrac{5}{36}\left(mol\right)\\ \Rightarrow V_{H_2\left(đktc\right)}=\dfrac{5}{36}.22,4=\dfrac{28}{9}\left(l\right)\)

Tóm tắt:

m_Al= 2,5(g)

\(n_{H_2SO_4}=0,5\left(mol\right)\)

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a) PTHH?

b) Chất dư? m?

c) Thể tích khí sau phản ứng?

Anh ít tóm tắt lắm nên không biết tóm tắt như ri ổn không, vì em cần nên anh tóm tắt ra thử

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!

\(n_{Zn}=\dfrac{26}{65}=0.4\left(mol\right)\)

\(Zn+2HCl\rightarrow ZnCl_2+H_2\)

\(0.4.......0.8....................0.4\)

\(m_{HCl}=0.8\cdot36.5=29.2\left(g\right)\)

\(V_{H_2}=0.4\cdot22.4=8.96\left(l\right)\)

\(nZn=\dfrac{13}{65}=0,2\left(mol\right)\)

\(Zn+H_2SO_4\rightarrow ZnSO_4+H_2\)

 1           1              1           1    (mol)

0,2       0,2             0,2        0,2    (mol)

\(VH_2=0,2.22,4=4,48\left(l\right)\)

\(Fe_2O_3+3H_2\rightarrow2Fe+3H_2O\)

 1             3         2             3    (mol)

              0,2         2/15            (mol)

\(mFe=\dfrac{2}{15}.56=7,47\left(g\right)\)

Oke

\(n_{HCl}=0,2.2=0,4\left(mol\right)\\ a.Zn+2HCl\rightarrow ZnCl_2+H_2\\ b.0,2........0,4.......0,2.......0,2\left(mol\right)\\ m=m_{Zn}=0,2.65=13\left(g\right)\\ c.m_{ZnCl_2}=136.0,2=27,2\left(g\right)\\ d.V_{H_2\left(đktc\right)}=0,2.22,4=4,48\left(l\right)\)

a)    Fe + H₂SO₄ → FeSO₄ + H₂

_{b)  Ta có :   nH2 = \(\dfrac{16,8}{22,4}\)} = 0,75 (mol)

⇒ n_Fe= 0,75.56 = 42(gam)

Ta có: \(n_{H_2}=\dfrac{16,8}{22,4}=0,75\left(mol\right)\)

a. \(PTHH:Fe+H_2SO_4--->FeSO_4+H_2\uparrow\)

b. Theo PT: \(n_{Fe}=n_{H_2}=0,75\left(mol\right)\)

\(\Rightarrow m_{Fe}=56.0,75=42\left(g\right)\)

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

Làm gộp các phần còn lại

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

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

\(n_{Fe}=\dfrac{28}{56}=0,5mol\)

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

0,5    1                           0,5

\(V_{H_2}=0,5\cdot22,4=11,2l\)

\(m_{HCl}=1\cdot36,5=36,5g\)