\(Fe + H_2SO_4 \to FeSO_4 + H_2\)

Chất rắn A : Cu

Theo PTHH : 

\(n_{Fe} = n_{H_2} = \dfrac{3,36}{22,4} = 0,15(mol)\\ \Rightarrow m_{Fe} = 0,15.56 = 8,4(gam)\)

Suy ra : \(m_{Cu} = 1,48 - 8,4 = -6,92 < 0\)→ Sai đề

PTHH: \(Fe+H_2SO_{4\left(l\right)}\rightarrow FeSO_4+H_2\uparrow\)

Ta có: \(\left\{{}\begin{matrix}n_{Fe}=\dfrac{22,4}{56}=0,4\left(mol\right)\\n_{H_2SO_4}=\dfrac{24,5}{98}=0,25\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\) Sắt còn dư, Axit p/ứ hết

\(\Rightarrow\left\{{}\begin{matrix}n_{H_2}=0,25\left(mol\right)\\n_{Fe\left(dư\right)}=0,15\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}V_{H_2}=0,25\cdot22,4=5,6\left(l\right)\\m_{Fe\left(dư\right)}=0,15\cdot56=8,4\left(g\right)\end{matrix}\right.\)

a, \(n_{Fe}=\dfrac{22,4}{56}=0,4\left(mol\right)\)

\(n_{H_2SO_4}=\dfrac{24,5}{98}=0,25\left(mol\right)\)

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

Xét tỉ lệ: \(\dfrac{0,4}{1}>\dfrac{0,25}{1}\), ta được Fe dư.

Theo PT: \(n_{Fe\left(pư\right)}=n_{H_2SO_4}=0,25\left(mol\right)\Rightarrow n_{Fe\left(dư\right)}=0,4-0,25=0,15\left(mol\right)\)

\(\Rightarrow m_{Fe\left(dư\right)}=0,15.56=8,4\left(g\right)\)

b, \(n_{H_2}=n_{H_2SO_4}=0,25\left(mol\right)\Rightarrow V_{H_2}=0,25.22,4=5,6\left(l\right)\)

Phương trình hóa học của phản ứng:

So sánh tỉ lệ  ⇒ Fe dư

Theo PT nFe (pư) = nH2SO4 = 0,25 mol ⇒ nFe dư = 0,4 – 0,25 = 0,15 mol

mFe dư = 0,15. 56 = 8,4g.

Do khối lượng Fe dư nên tính thể tích khí H2 theo số mol H2SO4.

nH2 = nH2SO4 = 0,25 mol

Vkhí = 0,25 . 22,4 = 5,6l.

a) nFe=0,4(mol); nH2SO4=0,5(mol)

PTHH: Fe + H2SO4 -> FeSO4 + H2

Ta có: 0,4/1 < 0,5/1

=> Fe hết, H2SO4 dư. tính theo nFe.

=> nH2= nH2SO4(p.ứ)=nFe=0,4(mol)

=>nH2SO4(dư)=0,5-0,4=0,1(mol)

=>H2SO4(dư)=0,1.98=9,8(g)

b) V(H2,dktc)=0,4.22,4=8,96(l)

PT: \(2Mg+O_2\underrightarrow{t^o}2MgO\)

Ta có: \(n_{Mg}=\dfrac{4,8}{24}=0,2\left(mol\right)\)

\(n_{O_2}=\dfrac{5,6}{22,4}=0,25\left(mol\right)\)

Xét tỉ lệ: \(\dfrac{0,2}{2}< \dfrac{0,25}{1}\), ta được O₂ dư.

Theo PT: \(n_{O_2\left(pư\right)}=\dfrac{1}{2}n_{Mg}=0,1\left(mol\right)\)

\(\Rightarrow n_{O_2\left(dư\right)}=0,25-0,1=0,15\left(mol\right)\)

Bạn tham khảo nhé!

PTHH: \(2Mg+O_2\underrightarrow{t^o}2MgO\)

Ta có: \(\left\{{}\begin{matrix}n_{Mg}=\dfrac{4,8}{24}=0,2\left(mol\right)\\n_{O_2}=\dfrac{5,6}{22,4}=0,25\left(mol\right)\end{matrix}\right.\)

Xét tỉ lệ: \(\dfrac{0,2}{2}< \dfrac{0,25}{1}\) \(\Rightarrow\) Magie p/ứ hết, Oxi còn dư

\(\Rightarrow n_{O_2\left(dư\right)}=0,25-0,1=0,15\left(mol\right)\)

\(a,n_{Al}=\dfrac{10,8}{27}=0,4\left(mol\right)\\ n_{O_2}=\dfrac{13,44}{22,4}=0,6\left(mol\right)\)

PTHH: 4Al + 3O₂ --t^o--> 2Al₂O₃

b, LTL: \(\dfrac{0,4}{4}>\dfrac{0,6}{3}\) => O₂ dư

Theo pthh: \(\left\{{}\begin{matrix}n_{O_2\left(pư\right)}=\dfrac{3}{4}n_{Al}=\dfrac{3}{4}.0,4=0,3\left(mol\right)\\n_{Al_2O_3}=\dfrac{1}{2}n_{Al}=\dfrac{1}{2}.0,4=0,2\left(mol\right)\end{matrix}\right.\)

=> V_{O2 (dư)} = (0,6 - 0,3).22,4 = 6,72 (l)

c, m_Al2O3 = 0,2.102 = 20,4 (g)

\(n_{Al}=\dfrac{10,8}{17}=0,4\left(mol\right)\\ n_{O_2}=\dfrac{13,44}{22,4}=0,6\left(mol\right)\\ pthh:4Al+3O_2\underrightarrow{t^o}2Al_2O_3\) 
\(LTL:\dfrac{0,4}{4}< \dfrac{0,6}{3}\) 
=> O2 dư P hết 
\(n_{Al_2O_3}=\dfrac{1}{2}n_{Al}=0,2\left(mol\right)\\ m_{Al_2O_3}=0,2.102=20,4g\)

\(n_{Al}=\dfrac{10,8}{27}=0,4mol\)

\(n_{O_2}=\dfrac{13,44}{22,4}=0,6mol\)

    \(4Al+3O_2\rightarrow\left(t^o\right)2Al_2O_3\)

Xét: \(\dfrac{0,4}{4}\) < \(\dfrac{0,6}{3}\)                  ( mol )

         0,4     0,3            0,2   ( mol )

Chất dư là O₂

\(m_{O_2\left(dư\right)}=\left(0,6-0,3\right).32=9,6g\)

\(m_{Al_2O_3}=0,2.102=20,4g\)

\(n_{Mg}=\dfrac{2.4}{24}=0.1\left(mol\right)\)

\(Mg+2HCl\rightarrow MgCl_2+H_2\)

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

\(V_{H_2}=0.1\cdot22.4=2.24\left(l\right)\)

\(2H_2+O_2\underrightarrow{^{t^0}}2H_2O\)

\(0.1.....0.05\)

\(m_{O_2\left(dư\right)}=\left(0.5-0.05\right)\cdot32=14.4\left(g\right)\)