Câu 29.

a)Dùng quỳ tím ẩm:

   +Hóa xanh: \(NaOH;Ca\left(OH\right)_2\)

     Nhỏ 1 lượng \(Na_2CO_3\) vào hai chất trên:

     Xuất hiện kết tủa: \(Ca\left(OH\right)_2\)

       \(Ca\left(OH\right)_2+Na_2CO_3\rightarrow CaCO_3\downarrow+2NaOH\)

     Không hiện tượng:\(NaOH\)

    +Qùy không đổi màu: \(NaCl;NaNO_3\)

      Nhỏ 1 ít bạc nitrat \(AgNO_3\) xuất hiện kết tủa: \(NaCl\)

       \(AgNO_3+NaCl\rightarrow AgCl\downarrow+NaNO_3\)

      Không hiện tượng:\(NaNO_3\)

Câu 30.

\(n_{\uparrow}=\dfrac{1,12}{22,4}=0,05mol\)

a)  \(ZnO+2HCl\rightarrow ZnCl_2+H_2O\)

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

     0,05     0,1                     0,05

\(m_{Zn}=0,05\cdot65=3,25\left(g\right)\)\(\Rightarrow\%m_{Zn}=\dfrac{3,25}{30}\cdot100\%=10,83\%\)

\(\Rightarrow m_{ZnO}=100\%-10,83\%=89,17\%\)

b)\(m_{ZnO}=30-3,25=26,75\left(g\right)\)

   \(\Rightarrow n_{ZnO}=\dfrac{26,75}{81}=0,33mol\)

  \(\Sigma n_{HCl}=0,33\cdot2+0,05\cdot2=0,76mol\)

  \(\Rightarrow m_{HCl}=0,76\cdot36,5=27,74\left(g\right)\)

 \(\Rightarrow m_{ddsau}=\dfrac{27,74}{7,4}\cdot100=374,86\left(g\right)\)

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

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

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

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

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

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

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

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

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

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

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

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

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)

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

Gọi số mol Zn, Al là a, b

=> 65a + 27b = 18,4 (1)

PTHH: Zn + 2HCl --> ZnCl₂ + H₂

           a----->2a------------->a

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

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

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

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

=> \(\left\{{}\begin{matrix}m_{Zn}=0,2.65=13\left(g\right)\\m_{Al}=0,2.27=5,4\left(g\right)\end{matrix}\right.\)

b) n_HCl(pư) = 2a + 3b = 1 (mol)

n_HCl(dư) = 0,6.2 - 1 = 0,2 (mol)

PTHH: KOH + HCl --> KCl + H₂O

           0,2<----0,2

=> \(V=\dfrac{0,2}{1}=0,2\left(l\right)\)

\(n_{HCl}=0,6.2=1,2\left(mol\right)\\ n_{H_2}=\dfrac{11,2}{22,4}=0,5\left(mol\right)\\ a,n_{HCl\left(dư\right)}=1,2-2.n_{H_2}=1,2-2.0,5=0,2\left(mol\right)\\PTHH:Zn+2HCl\rightarrow ZnCl_2+H_2\\ 2Al+6HCl\rightarrow2AlCl_3+3H_2\\ Đặt:n_{Zn}=a\left(mol\right);n_{Al}=b\left(mol\right)\left(a,b>0\right)\\ \Rightarrow\left\{{}\begin{matrix}65a+27b=18,4\\a+1,5b=0,5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=0,2\\b=0,2\end{matrix}\right.\\ \Rightarrow m_{Zn}=0,2.65=13\left(g\right);m_{Al}=0,2.27=5,4\left(g\right)\\ b,KOH+HCl_{dư}\rightarrow KCl+H_2O\\ n_{KOH}=n_{HCl\left(dư\right)}=0,2\left(mol\right)\\ \Rightarrow V=V_{ddKOH}=\dfrac{0,2}{1}=0,2\left(l\right)\)

a)\(\left\{{}\begin{matrix}n_{Zn}=x\left(mol\right)\\n_{Al}=y\left(mol\right)\end{matrix}\right.\)

\(n_{HCl}=V\cdot C_M=2\cdot0,6=1,2mol\)

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

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

 x         2x           x            x

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

 y         3y           y               1,5y

Ta có hệ: \(\left\{{}\begin{matrix}65x+27y=18,4\\x+1,5y=0,5\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0,2\\y=0,2\end{matrix}\right.\)

\(m_{Zn}=0,2\cdot65=13g\)

\(m_{Al}=0,2\cdot27=5,4g\)

b)Theo pt: \(\Sigma n_{HCl}=2x+3y=2\cdot0,2+3\cdot0,2=1mol\)

Để trung hòa: \(n_{OH^-}=n_{H^+}=1mol\)

\(\Rightarrow V=\dfrac{n_{OH^-}}{C_M}=\dfrac{1}{1}=1l\)

Ta có: \(n_{HCl}=0,4\cdot1,5=0,6\left(mol\right)\)

Bảo toàn Hidro: \(n_{H_2}=\dfrac{1}{2}n_{HCl}=0,3\left(mol\right)\) \(\Rightarrow V_{H_2}=0,3\cdot22,4=6,72\left(l\right)\)

\(\Rightarrow\) Chọn B