a) Fe + H2SO4 -----------> FeSO4 + H2

\(n_{Fe}=n_{H_2}=0,75\left(mol\right)\)

=> \(m_{Fe}=0,75.56=42\left(g\right)\)

b) \(CM_{H_2SO_4}=\dfrac{0,75}{0,25}=3M\)

c) \(m_{ddsaupu}=42+250.1,1-0,75.2=315,5\left(g\right)\)

=> \(C\%_{FeSO_4}=\dfrac{0,75.152}{315,5}.100=36,13\%\)

\(n_{SO_2}=\dfrac{16,8}{22,4}=0,75\left(mol\right)\)

\(PTHH:2Fe+6H_2SO_4\rightarrow Fe_2\left(SO_4\right)_3+3SO_2+6H_2O\)

 Mol:         0,5           1,5                   0,25          0,75       1,5

a)m_Fe=0,5.56=28 (g)

b)\(C_{MddH_2SO_4}=\dfrac{1,5}{0,25}=6\left(mol/l\right)\)

c)\(m_{Fe_2\left(SO_4\right)_3}=0,25.400=100\left(g\right)\)

  \(m_{H_2O}=1,5.18=27\left(g\right)\)

\(C\%_{ddFe_2\left(SO_4\right)_3}=\dfrac{100.100}{100+27}=78,74\%\)

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, \(n_{H_2SO_4}=n_{Fe}=0,35\left(mol\right)\Rightarrow C_{M_{H_2SO_4}}=\dfrac{0,35}{0,2}=1,75\left(M\right)\)

d, \(n_{FeSO_4}=n_{Fe}=0,35\left(mol\right)\Rightarrow m_{FeSO_4}=0,35.152=53,2\left(g\right)\)

e, \(C_{M_{FeSO_4}}=\dfrac{0,35}{0,2}=1,75\left(M\right)\)

d, \(n_{H_2SO_4}=0,25.1,6=0,4\left(mol\right)\)

Xét tỉ lệ: \(\dfrac{n_{Fe}}{1}< \dfrac{n_{H_2SO_4}}{1}\), ta được H₂SO₄ dư.

Theo PT: \(n_{H_2SO_4\left(pư\right)}=n_{Fe}=0,35\left(mol\right)\)

\(\Rightarrow n_{H_2SO_4\left(dư\right)}=0,4-0,35=0,05\left(mol\right)\)

\(\Rightarrow m_{H_2SO_4\left(dư\right)}=0,05.98=4,9\left(g\right)\)

Bài 1:

A . PTHH : 2AI + 3H₂SO₄ => AI₂(SO₄)₃ + 3H₂

B . nAI = 7.1 : 27 = 0.2 (mol)

 PT :2AI         +       3H₂SO₄      =>        AI₂(SO₄)₃       +   3H₂

        _{2 mol              3 mol                                                       3 mol}

        _{0.2 mol            0.3 mol                                                    0.3 mol}

=> V_{H2 = 22.4 X 0.3 = 6.72 (lít)}  

3.  500ml = 0.5 lít

Nồng độ mol/l của dd H₂SO₄ là:

C_{M H2SO4} = mol/lít =0.3/0.5 = 0.6 MOL/ LÍT

chuẩn nha hoàng kiều anh

a) 

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

b)  \(n_{Zn=\frac{16,12}{65}=0,248\left(mol\right)}\)

Theo PTHH :  \(n_{H_2}=n_{Zn}=0,248\left(mol\right)\)

\(V_{H_2\left(đktc\right)}=0,248\times22,4=5,5552\left(l\right)\)

c ) Ta có :

Theo PTHH :  \(n_{HCl}=2n_{Zn}=0,496\left(mol\right)\)

\(\Rightarrow m_{HCl}=18,104\left(g\right)\)

\(m_{dd}=\frac{18,104\times100}{10}=181,04\left(g\right)\)

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

b, Ta có: \(n_{H_2}=\dfrac{8,96}{22,4}=0,4\left(mol\right)\)

Theo PT: \(n_{Al}=\dfrac{2}{3}n_{H_2}=\dfrac{4}{15}\left(mol\right)\)

\(\Rightarrow\left\{{}\begin{matrix}\%m_{Al}=\dfrac{\dfrac{4}{15}.27}{10}.100\%=72\%\\\%m_{Cu}=28\%\end{matrix}\right.\)

c, Theo PT: \(n_{H_2SO_4}=n_{H_2}=0,4\left(mol\right)\)

\(\Rightarrow m_{H_2SO_4}=0,4.98=39,2\left(g\right)\)

\(\Rightarrow C\%_{H_2SO_4}=\dfrac{39,2}{300}.100\%\approx13,067\%\)

\(n_{H_2}=\dfrac{3,36}{22,4}0,15(mol)\\ a,PTHH:Fe+H_2SO_4\to FeSO_4+H_2\\ b,n_{Fe}=n_{H_2}=0,15(mol)\\ \Rightarrow \%_{Fe}=\dfrac{0,15.56}{14,8}.100\%=56,76\%\\ \Rightarrow \%_{Cu}=100\%-56,76\%=43,24\%\\ c,n_{H_2SO_4}=0,15(mol)\\ \Rightarrow m_{dd_{H_2SO_4}}=\dfrac{0,15.98}{20\%}=73,5(g)\\ \Rightarrow V_{dd_{H_2SO_4}}=\dfrac{73,5}{1,4}=52,5(l)\)

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

a) PTHH : \(Fe+H_2SO_4\rightarrow FeSO_4+H_2\)

Theo Pt : \(n_{Fe}=n_{H2SO4}=n_{FeSO4}=n_{H2}=0,2\left(mol\right)\)

b) \(V_{H2\left(dktc\right)}=0,2.22,4=4,48\left(l\right)\)

c) \(C_{MddH2SO4}=\dfrac{0,2}{0,2}=1\left(M\right)\)

d) \(m_{muối}=m_{FeSO4}=0,2.152=30,4\left(g\right)\)

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

PTHH: 2Al + 3H₂SO₄ --> Al₂(SO₄)₃ + 3H₂

           0,4-->0,6---------->0,2------->0,6

=> \(C_{M\left(dd.H_2SO_4\right)}=\dfrac{0,6}{0,15}=4M\)

b) V_H2 = 0,6.22,4 = 13,44 (l)

c) \(C_{M\left(Al_2\left(SO_4\right)_3\right)}=\dfrac{0,2}{0,15}=\dfrac{4}{3}M\)

\(n_{Al}=\dfrac{10,8}{27}=0,4\left(mol\right)\\ pthh:2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\) 
           0,4     0,6                   0,2             0,6 
\(C_M_{H_2SO_4}=\dfrac{0,6}{0,15}=4M\\ V_{H_2}=0,622,4=13,44L\) 
\(C_M=\dfrac{0,2}{0,15}=1,3M\)

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