Gọi x, y là số mol CuO, ZnO

CuO + 2HCl -----> CuCl2 + H2O

ZnO +  2HCl -----> ZnCl2 + H2O

Ta có :\(\left\{{}\begin{matrix}80x+81y=12,1\\2x+2y=0,4.3=1,2\end{matrix}\right.\)

=> x=36,5 ; y=-35,9 

Xem lại đề !

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

PTHH: 2Al + 6HCl → 2AlCl₃ + 3H₂

Mol:       x                                  1,5x

PTHH: Zn + 2HCl → ZnCl₂ + H₂

Mol:       y                                 y

Ta có: \(\left\{{}\begin{matrix}27x+65y=24,9\\1,5x+y=0,6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}27x+65.\left(0,6-1,5x\right)=24,9\\y=0,6-1,5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0,2\\y=0,3\end{matrix}\right.\)

b, \(m_{Al}=0,2.27=5,4\left(g\right);m_{Zn}=24,9-5,4=19,5\left(g\right)\)

c) \(\%m_{Al}=\dfrac{5,4.100\%}{24,9}=21,69\%;\%m_{Zn}=100\%-21,69\%=78,31\%\)

d) 

PTHH: 2Al + 6HCl → 2AlCl₃ + 3H₂

Mol:      0,2     0,6          0,2       0,3

PTHH: Zn + 2HCl → ZnCl₂ + H₂

Mol:      0,3     0,6        0,3       0,3

\(m_{ddHCl}=\dfrac{\left(0,6+0,6\right).36,5.100}{14}=312,857\left(g\right)\)

e) m_{dd sau pứ} = 24,9 + 312,857 - (0,3+0,3).2 = 336,557 (g)

\(C\%_{ddAlCl_3}=\dfrac{0,2.133,5.100\%}{336,557}=7,93\%\)

\(C\%_{ddZnCl_2}=\dfrac{0,3.136.100\%}{336,557}=12,12\%\)

\(n_{H_2}=\dfrac{0.56}{22.4}=0.025\left(mol\right)\)

\(2Al+6HCl\rightarrow2AlCl_3+3H_2\)

\(n_{Al}=\dfrac{2}{3}\cdot0.025=\dfrac{1}{60}\left(mol\right)\)

\(m_{Al}=\dfrac{1}{60}\cdot27=0.45\left(g\right)\)

\(m_{Cu}=25-0.45=24.55\left(g\right)\)

\(\%Cu=\dfrac{24.55}{25}\cdot100\%=98.2\%\)

\(\%Al=100-98.2=1.8\%\)

\(Cu+2H_2SO_{4\left(đ\right)}\rightarrow CuSO_4+SO_2+2H_2O\)

\(2Al+6H_2SO_{4\left(đ\right)}\rightarrow Al_2\left(SO_4\right)_3+3SO_2+6H_2O\)

\(n_{Cu}=\dfrac{24.55}{64}=\dfrac{491}{1280}\left(mol\right)\)

\(V_{SO_2}=\left(\dfrac{1}{60}\cdot\dfrac{3}{2}+\dfrac{491}{1280}\right)\cdot22.4=9.1525\left(l\right)\)

a)

$2Al + 6HCl \to 2AlCl_3 + 3H_2$
b)

$n_{H_2} = \dfrac{0,56}{22,4} = 0,225(mol)$

Theo PTHH : $n_{Al} = \dfrac{2}{3}n_{H_2} = \dfrac{1}{60}(mol)$
$m_{Al} = \dfrac{1}{60}.27 = 0,45(gam)$
$m_{Cu} = 25 - 0,45 = 24,55(gam)$

c)

$\%m_{Al} = \dfrac{0,45}{25}.100\% = 1,8\%$
$\%m_{Cu} = 100\% -1,8\% = 98,2\%$

d)

$Cu + 2H_2SO_4 \to CuSO_4 + SO_2 + 2H_2O$
$2Al + 6H_2SO_4 \to Al_2(SO_4)_3 + 3SO_2 + 6H_2O$

Theo PTHH : 

$n_{SO_2} = n_{Cu} + \dfrac{3}{2}n_{Al} = \dfrac{24,55}{64} + \dfrac{1}{60}.\dfrac{3}{2} = 0,41(mol)$
$V_{SO_2} = 0,41.22,4 = 9,184(lít)$

2Al+6HCl->2AlCl3+ 3H2

n(H2)=0,025mol

n(al)= 1/60 mol

m(Al)= 0,45g

m(Cu)= 25-0,45=24,55g

%m(Cu)=98,2%

%m(Al)=1,8%

Cu+2H2SO4(đ)-> CuSO4+SO2+H2O

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

n(so2)=n(Cu)+1,5n(al)= 24,55/64 +1,5*1/60 =523/1280 mol

V(so2)=9,1525lit

2Al + Fe2O3 -> Al2O3 + 2Fe

             0.1        0.1

Al2O3 + 2NaOH -> H2O + 2NaAlO2

  0.1                                       0.2

n(NaAlO2) = 14.8/82=0.2 mol

m(Fe2O3)= 0.1*160=16(g) ->m Al =25 - 16 = 9(g)

%m(Al)=(9*100)/25=36 %

giup mk voi hay cho biet giới han đo cua thươc cuon

la50%

Ta có: \(n_{HCl}=2.\dfrac{200}{1000}=0,4\left(mol\right)\)

PTHH:

\(Fe+2HCl--->FeCl_2+H_2\uparrow\left(1\right)\)

\(MO+2HCl--->MCl_2+H_2O\left(2\right)\)

Gọi x, y lần lượt là số mol của Fe và MO

Theo PT₍₁₎: \(n_{HCl}=2.n_{Fe}=2x\left(mol\right)\)

Theo PT₍₂₎: \(n_{HCl}=2.n_{MO}=2y\left(mol\right)\)

\(\Rightarrow2x+2y=0,4\) (*)

Theo đề, ta có: \(56x+My+16y=11,2\) (**)

Từ (*) và (**), ta có:

\(\left\{{}\begin{matrix}2x+2y=0,4\\56x+16y+My=11,2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}16x+16y=3,2\\56x+16y+My=11,2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}40x+My=8\\2x+2y=0,4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}40x+My=8\\40x+40y=8\end{matrix}\right.\Leftrightarrow M=40\)

Vậy M là nguyên tử canxi (Ca)

a) nH2=1,12/22,4=0,05(mol)

PTHH: Zn +2 HCl -> ZnCl2 + H2

0,05____0,1_____0,05____0,05(mol)

=>mZn= 0,05.65= 3,25(g)

b) => %mZn= (3,25/20).100=16,25%

=>%mZnO=100% - 16,25%= 83,75%

c) mZnO= 20 - 3,25= 16,75(g) => nZnO= 16,75/81= 67/324(mol)

PTHH: ZnO +2 HCl -> ZnCl2 + H2O

67/324______67/162(mol)

=> mddHCl= [(67/162+ 0,1). 36,5]: 14%=133,898(g)