2Cu+O₂-to>2CuO

0,1----0,05

4Al+3O₂-to>2Al₂O₃

0,4---0,3 mol

n _Cu=\(\dfrac{6,4}{64}\)=0,1 mol

n _Al=\(\dfrac{10,8}{27}\)=0,4 mol

=>V_O2=0,35.22,4=7,84l

2Cu+O₂-to>2CuO

0,1----0,05

4Al+3O₂-to>2Al₂O₃

0,4---0,3 mol

n _Cu=10,8/27=0,4 mol

=>V_O2=0,35.22,4=7,84l

Đốt 9,4 (g) h2 Mg; Na đủ với 3,36(l) O2 đktc. Xđ % m Mg

Xem lại đề bại ơi

nH2 = 4,48/22,4 = 0,2 (mol)

PTHH: Fe + 2HCl -> FeCl2 + H2

nFe = nH2 = 0,2 (mol)

mFe = 0,2 . 56 = 11,2 (g)

n02=0,1mol

PThh: 4Al+3O2=>2Al2O3

           2/15<-0,1->1/15

=> mAl2O3=1/15.102=6,8g

OK chứ ??

2Al + 3O₂ ----> 2Al₂O3

nO2=2,24/22,4=0.1mol

nAl2O3=nO2=0,1 mol

----> mAl2O3=0,1*102=10.2g

Gọi số mol Al, Zn là a, b (mol)

=> 27a + 65b = 11,9 (1)

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

PTHH: 2Al + 6HCl --> 2AlCl₃ + 3H₂

             a----------------------->1,5a

           Zn + 2HCl --> ZnCl₂ + H₂

           b------------------------>b

=> 1,5a + b = 0,4 (2)

(1)(2) => a = 0,2; b = 0,1

\(\left\{{}\begin{matrix}\%Al=\dfrac{0,2.27}{11,9}.100\%=45,38\%\\\%Zn=\dfrac{0,1.65}{11,9}.100\%=54,62\%\end{matrix}\right.\)

2Al+3H₂SO₄->Al₂(SO₄)₃+3H₂

x------------------------------------\(\dfrac{3}{2}x\)

Zn+H₂SO₄->ZnSO₄+H₂

y-----------------------------y

=>Ta có :\(\left\{{}\begin{matrix}27x+65y=11,9\\\dfrac{3}{2}x+y=0,4\end{matrix}\right.\)

=>x=0,2 mol , y=0,1 mol

=>%m _Al=\(\dfrac{0,2.27}{11,9}.100\)=45,38%

=>%m _Zn=54,62%

\(BTKL:\)

\(m_{O_2}=34-20.4=13.6\left(g\right)\)

\(n_{O_2}=\dfrac{13.6}{32}=0.425\left(mol\right)\)

\(BTNTO:\)

\(n_{H_2O}=2n_{O_2}=2\cdot0.425=0.85\left(mol\right)\)

\(BTNTH:\)

\(n_{HCl}=2n_{H_2O}=2\cdot0.85=1.7\left(mol\right)\)

\(m_{Muối}=m_{Kl}+m_{Cl^-}=20.4+1.7\cdot35.5=80.75\left(g\right)\)

nH2 = 13,44/22,4 = 0,6 (mol)

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

Mol: 0,4 <--- 0,6 <--- 0,2 <--- 0,6

mAl = 0,4 . 27 = 10,8 (g)

nO2 = 11,2/22,4 = 0,5 (mol)

PTHH: 2H2 + O2 -> (t°) 2H2O

LTL: 0,6/2 < 0,5 => O2 dư

VO2 (p/ư) = 13,44/2 = 6,72 (l)

VO2 (dư) = 11,2 - 6,72 = 4,48 (l)

nH2O = 0,6 (mol)

mH2O = 0,6 . 18 = 10,8 (g)

a.\(n_{H_2}=\dfrac{13,44}{22,4}=0,6mol\)

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

0,4                                             0,6    ( mol )

\(m_{Al}=0,4.27=10,8g\)

b.\(n_{O_2}=\dfrac{11,2}{22,4}=0,5mol\)

\(2H_2+O_2\rightarrow\left(t^o\right)2H_2O\)

0,6 <   0,5                         ( mol )

0,6      0,3             0,6           ( mol )

\(V_{O_2\left(dư\right)}=\left(0,5-0,3\right).22,4=4,48l\)

\(m_{H_2O}=0,6.18=10,8g\)

nH2 = \(\dfrac{13,44}{22,4}\)= 0,6 (mol)

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

Mol: 0,4 <--- 0,6 <--- 0,2 <--- 0,6

mAl = 0,4 . 27 = 10,8 (g)

nO2 = 11,2/22,4 = 0,5 (mol)

PTHH: 2H2 + O2 -to> 2H2O

=> O2 dư

VO2 pứ = 13,44/2 = 6,72 (l)

VO2 dư = 11,2 - 6,72 = 4,48 (l)

nH2O = 0,6 (mol)

mH2O = 0,6 . 18 = 10,8 (g)

Gọi số mol Fe, Zn là a, b (mol)

=> 56a + 65b = 12,1 (1)

\(n_{H_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\)

PTHH: Fe + 2HCl --> FeCl₂ + H₂

             a----------------------->a

            Zn + 2HCl --> ZnCl₂ + H₂

             b----------------------->b

=> a + b =0,2 (2)

(1)(2) => a = 0,1 (mol); b = 0,1 (mol)

=> \(\left\{{}\begin{matrix}\%m_{Fe}=\dfrac{0,1.56}{12,1}.100\%=46,28\%\\\%m_{Zn}=\dfrac{0,1.65}{12,1}.100\%=53,72\%\end{matrix}\right.\)