a) m_Cu = 3,2 (g)

=> m_Fe = 6 - 3,2 = 2,8 (g)

\(n_{Fe}=\dfrac{2,8}{56}=0,05\left(mol\right)\)

PTHH: Fe + 2HCl --> FeCl₂ + H₂

          0,05->0,1--->0,05--->0,05

=> V₁ = 0,05.22,4 = 1,12 (l)

\(n_{Cu}=\dfrac{3,2}{64}=0,05\left(mol\right)\)

PTHH: 2Fe + 6H₂SO_4(đ/n) --> Fe₂(SO₄)₃ + 3SO₂ + 6H₂O

          0,05--------------------------------->0,075

            Cu + 2H₂SO₄ --> CuSO₄ + SO₂ + 2H₂O

           0,05------------------------>0,05

=> V₂ = (0,075 + 0,05).22,4 = 2,8 (l)

b)

n_HCl(dư) = 0,5.2 - 0,1 = 0,9 (mol)

=> \(\left\{{}\begin{matrix}C_{M\left(HCl.dư\right)}=\dfrac{0,9}{0,5}=1,8M\\C_{M\left(FeCl_2\right)}=\dfrac{0,05}{0,5}=0,1M\end{matrix}\right.\)

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

\(Cu+2H_2SO_{4\left(đ\right)}\underrightarrow{t^o}CuSO_4+SO_2+2H_2O\)

b, Ta có: \(n_{H_2}=\dfrac{2,479}{24,79}=0,1\left(mol\right)\)

\(n_{Zn}=n_{H_2}=0,1\left(mol\right)\)

\(n_{SO_2}=\dfrac{2,9748}{24,79}=0,12\left(mol\right)\)

\(n_{Cu}=n_{SO_2}=0,12\left(mol\right)\)

\(\Rightarrow m=m_{Zn}+m_{Cu}=0,1.65+0,12.64=14,18\left(g\right)\)

Có: \(n_{H_2SO_{4\left(đ\right)}}=2n_{SO_2}=0,24\left(mol\right)\Rightarrow x=m_{ddH_2SO_4\left(đ\right)}=\dfrac{0,24.98}{98\%}=24\left(g\right)\)

a) \(n_{H_2}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\)

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

           0,1<----------------------------0,15

=> \(\%m_{Al}=\dfrac{0,1.27}{7,5}.100\%=36\%\)

\(\%m_{Cu}=100\%-36\%=64\%\)

b) \(n_{Cu}=\dfrac{7,5-0,1.27}{64}=0,075\left(mol\right)\)

PTHH: Cu + 2H₂SO₄ --> CuSO₄ + SO₂ + 2H₂O

          0,075------------------------>0,075

            2Al + 6H₂SO₄ --> Al₂(SO₄)₃ + 3SO₂ + 6H₂O

              0,1----------------------------->0,15

=> V_SO2 = (0,075 + 0,15).22,4 = 5,04 (l)

2Al+6HCl->2AlCL3+3H2(1)

Mg+2HCL->MgCL2+H2(2)

Cu+2H2SO4(đặc)->CuSO4+SO2+2H2O(3)

(3)=>nCu=nSO2=2,24/22,4=0,1mol

=>mCu=0,1.64=6,4g ???? lớn hơn m hỗn hợp đề chắc sai

Bài 1:

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

            \(Fe_2O_3+3H_2SO_4\rightarrow Fe_2\left(SO_4\right)_3+3H_2O\)

Ta có: \(n_{H_2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)=n_{Fe}\)

\(\Rightarrow\%m_{Fe}=\dfrac{0,1\cdot56}{37,6}\cdot100\%\approx14,89\%\)

\(\Rightarrow\%m_{Fe_2O_3}=85,11\%\) 

Bài 3: 

PTHH: \(2HNO_3+Ba\left(OH\right)_2\rightarrow Ba\left(NO_3\right)_2+2H_2O\)

Ta có: \(\left\{{}\begin{matrix}n_{HNO_3}=0,05\cdot1=0,05\left(mol\right)\\n_{Ba\left(OH\right)_2}=\dfrac{342\cdot5\%}{171}=0,1\left(mol\right)\end{matrix}\right.\)

Xét tỉ lệ: \(\dfrac{0,05}{2}< \dfrac{0,1}{1}\) \(\Rightarrow\) Axit p/ứ hết, Bazơ còn dư sau p/ứ

\(\Rightarrow\) Dung dịch sau p/ứ làm quỳ tím hóa xanh

Theo PTHH: \(n_{Ba\left(NO_3\right)_2}=\dfrac{1}{2}n_{HNO_3}=0,025\left(mol\right)\) \(\Rightarrow m_{Ba\left(NO_3\right)_2}=0,025\cdot261=6,525\left(g\right)\)

Bài 2:

Theo đề bài: \(\%m_{Cu}=\dfrac{6,4}{14,2}\cdot100\%\approx45,07\%\)

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

               a_________________________\(\dfrac{3}{2}\)a   (mol)

            \(Mg+H_2SO_4\rightarrow MgSO_4+H_2\uparrow\)

                b_____________________b     (mol)

Ta lập được hệ phương trình \(\left\{{}\begin{matrix}27a+24b=14,2-6,4=7,8\\\dfrac{3}{2}a+b=\dfrac{8,96}{22,4}=0,4\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}a=0,2\\b=0,1\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}\%m_{Al}=\dfrac{0,2\cdot27}{14,2}\cdot100\%\approx38,03\%\\\%m_{Mg}=100\%-38,03\%-45,07\%=16,9\%\end{matrix}\right.\)

Qui đổi ½ hh B gồm Al (x mol), Fe (y mol), O (z mol)

=> mB = 2 (mAl + mFe + mO) = 102,78g

Gọi công thức của oxit sắt là Fe_aO_b

=> Fe₂O₃

n khí = 0,6 mol

nNO    30                5,34

                  40,66

nNO2   46              10,66

=> nNO / nNO2 = 1/2  

=> nNO = 0,2   nNO2 = 0,4

5Cu +   16HNO3 =>   5Cu(NO3)2   + 2 NO +   4NO2   +    8H2O

0,5                                                 <= 0,2

=> m = 0,5.64 = 32 g