a) n_NO=8,96/22,4=0,4(mol)

bảo toàn e:

Al⁰ --> Al³⁺ +3e

x                    3x

Cu⁰--> cu⁺²+ + 2e

  y                      2y

N⁺⁵ + 3e --> N⁺²

           1,2       0,4

=> hệ: 27x+64y=24,6 và 3x+2y=1,2

bấm máy ta được: x= 0,2 và y=0,3

=> m_Al=5,4g ; m_Cu=19,2g

b) Al + 4HNO₃ --> Al(NO₃)₃ + NO + 2H₂O

    3Cu + 8HNO₃ --> 3Cu(NO₃)₂ + 2NO + 4H₂O 

Ta thấy: n_{HNO3(phản ứng)}=4n_NO=1,6(mol)

=> n_{HNO3(đã dùng)}=1,6+1,6*20%=1,92(mol)

=>V_{HNO3(đã dùng)}=1,92/1=1,92(l)

a, Gọi \(\left\{{}\begin{matrix}n_{Fe}=a\left(mol\right)\\n_{Al}=b\left(mol\right)\end{matrix}\right.\)

\(n_{H_2}=\dfrac{11,2}{22,4}=0,5\left(mol\right)\)

PTHH:

Fe + 2HCl ---> FeCl₂ + H₂

a--->2a------------------>a

2Al + 6HCl ---> 2AlCl₃ + 3H₂

b---->3b-------------------->1,5b

=> \(\left\{{}\begin{matrix}56a+27b=16,6\\a+1,5b=0,5\end{matrix}\right.\Leftrightarrow a=b=0,2\left(mol\right)\)

=> \(\left\{{}\begin{matrix}m_{Fe}=0,2.56=11,2\left(g\right)\\m_{Al}=0,2.27=5,4\left(g\right)\end{matrix}\right.\)

b) \(C\%_{HCl}=\dfrac{\left(0,2.2+0,2.3\right).36,5}{300}.100\%=12,167\%\)

\(n_{H_2}=\dfrac{11,2}{22,4}=0,5\left(mol\right)\) 
gọi nFe : a , nAl: b (a,b>0)  => 56a + 27b = 16,6 (g) 
\(pthh:Fe+2HCl\rightarrow FeCl_2+H_2\uparrow\) 
           a                                     a
         \(2Al+6HCl\rightarrow2AlCl_3+3H_2\uparrow\) 
           b                                     \(\dfrac{3b}{2}\)
          
=> \(a+\dfrac{3b}{2}=0,5\) 
ta có hệ pt 
\(\left\{{}\begin{matrix}56a+27b=16,6\\a+\dfrac{3b}{2}=0,5\end{matrix}\right.\) 
=> a= 0,2 , b = 0,2 
\(\left\{{}\begin{matrix}m_{Fe}=0,2.56=11,2\left(g\right)\\m_{Al}=16,6-11,2=5,4\left(g\right)\end{matrix}\right.\) 
\(pthh:Fe+2HCl\rightarrow FeCl_2+H_2\) 
           0,2     0,4 
        \(2Al+6HCl\rightarrow2AlCl_3+3H_2\) 
           0,2    0,6 
=> \(m_{HCl}=\left(0,4+0,6\right).36,5=36,5\left(g\right)\) 
=> \(C\%=\dfrac{36,5}{200}.100\%=18,25\%\)

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

PTHH: Ba + 2H₂O --> Ba(OH)₂ + H₂

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

=> m_Ba = 0,1.137 = 13,7 (g)

=> m_Cu = 20 - 13,7 = 6,3 (g)

\(\left\{{}\begin{matrix}\%m_{Ba}=\dfrac{13,7}{20}.100\%=68,5\%\\\%m_{Cu}=\dfrac{6,3}{20}.100\%=31,5\%\end{matrix}\right.\)

Ba+2H₂O->Ba(OH)₂+H₂

0,1--------------------------0,1 mol

n _H2 =\(\dfrac{2,24}{22,4}=0,1mol\)

m _Ba=0,1.137=13,7g=>%Ba=68,5%

=>m _Cu=20-13,7=6,3g=>%Cu=31,5%

Vik chỉ có Ba tác dụng đc vs nước (H₂O) , Cu ko td vs nước nên ta có :

PTHH :  \(Ba+2H_2O->Ba\left(OH\right)_2+H_2\uparrow\)       (1)

\(n_{H_2}=\dfrac{V_{\left(đktc\right)}}{22,4}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\)

Từ (1) ->  \(n_{Ba}=n_{H_2}=0,1\left(mol\right)\)

-> \(m_{Ba}=n.M=0,1.137=13,7\left(g\right)\)

=> \(\%m_{Ba}=\dfrac{m_{Ba}}{m_{HH}}.100\%=68,5\%\)

\(\%m_{Cu}=100\%-\%m_{Ba}=100\%-68,5\%=31,5\%\)

1.1. Al + NaOH + H2O ==> NaAlO2 + 3/2H2

nH2(1)=3,36/22,4=0.15(mol)

=> nAl(1)= nH2(1):3/2= 0.15:3/2= 0.1(mol)

2.Mg + 2HCl ==> MgCl2 + H2

3.2Al + 6HCl ==> 2AlCl3 + 3H2

4.Fe + 2HCl ==> FeCl2 + H2

=> \(n_{H_2\left(2,3,4\right)}=\) 10.08/22.4= 0.45(mol)

=> nH2(3)=0.1*3/2=0.15(mol)

MgCl2 + 2NaOH ==> Mg(OH)2 + 2NaCl

AlCl3 + 3NaOH ==> Al(OH)3 + 3NaCl

FeCl2 + 2NaOH ==> Fe(OH)2 + 2NaCl

Mg + 2 H2SO4 (đ) -to-> MgSO4 + SO2 + 2 H2O

x_________2x__________________x(mol)

2 Fe + 6 H2SO4(đ) -to-> Fe2(SO4)3 + 3 SO2 + 6 H2O

y______3y_____________________1,5y(mol)

Ta có hpt:

\(\left\{{}\begin{matrix}24x+56y=18,4\\x+1,5y=0,6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0,3\\y=0,2\end{matrix}\right.\)

=> mMg= 0,3.24=7,2(g)

=> %mMg= (7,2/18,4).100=39,13%

=>%mFe= 60,87%

b) nH2SO4(tổng)=2x+3y=2.0,3+3.0,2=1,2(mol)

VddH2SO4=1,2/2=0,6(l)

a) Đặt \(\left\{{}\begin{matrix}n_{Fe}=a\left(mol\right)\\n_{Mg}=b\left(mol\right)\end{matrix}\right.\) \(\Rightarrow56a+24b=18,4\)  (1)

Ta có: \(n_{SO_2}=\dfrac{13,44}{22,4}=0,6\left(mol\right)\)

Bảo toàn electron: \(3a+2b=0,6\cdot2\)  (2)

Từ (1) và (2) \(\Rightarrow\left\{{}\begin{matrix}a=0,2\\b=0,3\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}\%m_{Fe}=\dfrac{0,2\cdot56}{18,4}\cdot100\%\approx60,87\%\\\%m_{Mg}=39,13\%\end{matrix}\right.\)

b) Ta có: \(\left\{{}\begin{matrix}n_{Fe_2\left(SO_4\right)_3}=\dfrac{1}{2}n_{Fe}=0,1\left(mol\right)\\n_{MgSO_4}=n_{Mg}=0,3\left(mol\right)\end{matrix}\right.\)

Bảo toàn nguyên tố: \(n_{H_2SO_4}=n_{SO_2}+3n_{Fe_2\left(SO_4\right)_3}+n_{MgSO_4}=1,2\left(mol\right)\)

\(\Rightarrow V_{H_2SO_4}=\dfrac{1,2}{2}=0,6\left(l\right)=600\left(ml\right)\)

a) Gọi mol Fe là a
           mol Mg là b
Ta có: 56a+24b = 18,4 (1)
Bte: 3a+2b = nS0₂.2
\(\Leftrightarrow\)3a+2b = 0,6.2  (2)
Từ(1),(2)\(\Rightarrow\left\{{}\begin{matrix}a=0,2\\b=0,3\end{matrix}\right.\)
\(\Rightarrow\)mFe = 11,2 (g) \(\Rightarrow\)%mFe = 60,87%
    mMg = 7,2 (g)  \(\Rightarrow\) %mMg = 39,13%
b)Ta có: nH₂S0₄ = 2nSo₂+nO_2(oxit)+nC0_{2(nếu có)}
\(\Rightarrow\)nH₂SO₄ = 2.0,6 = 1,2 (mol)
Có: C_M = \(\dfrac{n}{V}\)
\(\Rightarrow\)V = n.C_M = 1,2.2 = 2,4 (\(l\))

 

\(n_{H_2}=\dfrac{0,56}{22,4}=0,025\left(mol\right)\)

\(PTHH:Ba+2H_2O\rightarrow Ba\left(OH\right)_2+H_2\uparrow\)

            0,025                                     0,025

\(BaO+H_2O\rightarrow Ba\left(OH\right)_2\)

  \(\rightarrow m_{Ba}=0,025.137=3,425\left(g\right)\\ \rightarrow\left\{{}\begin{matrix}\%m_{Ba}=\dfrac{3,425}{6,486}=52,81\%\\\%m_{BaO}=100\%-52,81\%=47,19\%\end{matrix}\right.\)

n SO2 = 1,568/22,4 = 0,07(mol)

Trong bài toán kim loại tác dụng với H2SO4, ta luôn có :

n H2SO4 pư = 2n SO2 = 0,07.2 = 0,14(mol)

=> m dd H2SO4 = 0,14.98/98% = 14(gam)

Đáp án B

Các phương trình hóa học:

M_xO_y + yCO → xM + yCO₂

Ca(OH)₂ + CO₂ → CaCO₃ + H₂O

Ta có: m_oxit = m_{kim loại} + m_oxi

Trong đó:

→m_oxit = 2,5 + 0,15.16 = 4,9 gam