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a)
n Fe = a(mol) ; n Zn = b(mol)
Fe + 2HCl → FeCl2 + H2
a.........2a...........a...................(mol)
Zn + 2HCl → ZnCl2 + H2
b.........2b.............b.........................(mol)
n HCl = 2a + 2b = 0,5.0,4 = 0,2(mol)
m muối = 127a + 136b = 10,52(gam)
=> a = 0,342 ; b = -0,243 < 0
=> Sai đề
Fe + 2 HCl -> FeCl2+ H2
x____2x____x_____x(mol)
Zn +2 HCl -> ZnCl2 + H2
y____2y____y______y(mol)
Ta có: nHCl(tổng)= 0,2 (mol)
<=> 2x+2y=0,2 (1)
Mặt khác: m(muối)= 10,52(g)
<=> 127x+ 136y=10,52 (2)
Từ (1), (2) ta có hpt: \(\left\{{}\begin{matrix}2x+2y=0,2\\127x+136y=10,52\end{matrix}\right.\)
Bấm ra có nghiệm âm??
a) Gọi n Zn = a(mol) ; n ZnO = b(mol)
=> 65a + 81b = 14,6(1)
$Zn + 2HCl \to ZnCl_2 + H_2$
$ZnO + 2HCl \to ZnCl_2 + H_2O$
n ZnCl2 = a + b = 27,2/136 = 0,2(2)
Từ (1)(2) suy ra : a = b = 0,1
%m Zn = 0,1.65/14,6 .100% = 44,52%
%m ZnO = 100% -44,52% = 55,45%
b)
n HCl = 2n Zn + 2n ZnO = 0,4(mol)
m dd HCl = 0,4.36,5/7,3% = 200(gam)
Gọi \(\left\{{}\begin{matrix}n_{Al}=x\left(mol\right)\\n_{Fe}=y\left(mol\right)\end{matrix}\right.\)
\(n_{H_2}=\dfrac{4,48}{22,4}=0,2mol\)
\(2Al+6HCl\rightarrow2AlCl_3+3H_2\)
\(Fe+2HCl\rightarrow FeCl_2+H_2\)
Theo pt: \(\Rightarrow\left\{{}\begin{matrix}3x+y=0,2\\27x+56y=5,5\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=\dfrac{19}{470}\\y=\dfrac{37}{470}\end{matrix}\right.\)
\(\%m_{Al}=\dfrac{\dfrac{19}{470}\cdot27}{5,5}\cdot100\%=19,84\%\)
\(\%m_{Fe}=100\%-19,84\%=80,16\%\)
a, PT: \(Zn+2HCl\rightarrow ZnCl_2+H_2\)
\(Fe+2HCl\rightarrow FeCl_2+H_2\)
b, Ta có: \(n_{HCl}=0,5.0,2=0,1\left(mol\right)\)
\(n_{H_2}=\dfrac{0,896}{22,4}=0,04\left(mol\right)\)
Theo PT: \(n_{HCl\left(pư\right)}=2n_{H_2}=0,08\left(mol\right)< 0,1\left(mol\right)\)
→ HCl dư.
Gọi: \(\left\{{}\begin{matrix}n_{Zn}=x\left(mol\right)\\n_{Fe}=y\left(mol\right)\end{matrix}\right.\)
Theo PT: \(n_{H_2}=n_{Zn}+n_{Fe}=x+y=0,04\left(1\right)\)
\(\left\{{}\begin{matrix}n_{ZnCl_2}=n_{Zn}=x\left(mol\right)\\n_{FeCl_2}=n_{Fe}=y\left(mol\right)\end{matrix}\right.\)⇒ 136x + 127y = 5,26 (2)
Từ (1) và (2) \(\Rightarrow\left\{{}\begin{matrix}x=0,02\left(mol\right)\\y=0,02\left(mol\right)\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\%m_{Zn}=\dfrac{0,02.65}{0,02.65+0,02.56}.100\%\approx53,72\%\\\%m_{Fe}\approx46,28\%\end{matrix}\right.\)
a)
\(n_{H_2} = \dfrac{3,36}{22,4} = 0,15(mol)\\ n_{Mg} = a\ mol; n_{Fe} = b\ mol\\ Mg + 2HCl \to MgCl_2 + H_2\\ Fe + 2HCl \to FeCl_2 + H_2 \)
Theo PTHH, ta có:
\(\left\{{}\begin{matrix}24a+56b=5,2\\a+b=0,15\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}a=0,1\\b=0,05\end{matrix}\right.\)
Suy ra:
\(\%m_{Mg} = \dfrac{0,1.24}{5,2}.100\% = 46,15\%\\ \%m_{Fe} = 100\% - 46,15\% = 53,85\% \)
b)
\(n_{HCl} = 2n_{H_2} = 0,15.2 = 0,3(mol)\\ \Rightarrow V_{dd\ HCl} = \dfrac{0,3}{1} = 0,3(lít) \)
Đặt :
nMg = a mol
nFe= b mol
mhh = 24a + 56b = 5.2 (g) (1)
Mg + 2HCl => MgCl2 + H2
Fe + 2HCl => FeCl2 + H2
nH2 = a + b = 0.15 (2)
(1) , (2)
a = 0.1
b = 0.05
%Mg = 2.4/5.2 * 100% = 46.15%
%Fe = 100 - 46.15 = 53.85%
nHCl = 2a + 2b = 0.05 * 2 + 0.1*2 = 0.3 (mol)
VddHCl = 0.3/1=0.3 (l)
nHCl=5.10-3 mol
2Na + 2H2O --> 2NaOH + H2
x mol x mol 1/2 mol
Ba + 2H2O --> Ba(OH)2 + H2
y mol y mol y mol
NaOH + HCl --> NaCl + H2O
x mol x mol
Ba(OH)2 + 2HCl--> BaCl2 + H2O
y mol 2y mol
Ta duoc: 23x + 137y =0,297 (1)
x + 2y =5.10-3 (2)
Tu (1) va (2) ta duoc => x= 10-3
=> y= 2.10-3
a/ mNa= 10-3.23=0,023g
mBa=2.10-3.137=0,274g
b/ nH2= 10-6 mol
H2 + O2 --> H2O
10-6 mol 10-6 mol
VO2= 10-6. 22,4=2,24.10-5 lit
VKK= 2,24.10-5.100/20=1,12.10-4 lit
\(Fe+H_2SO_4\rightarrow FeSO_4+H_2\) (1)
\(Zn+H_2SO_4\rightarrow ZnSO_4+H_2\) (2)
Đặt \(n_{Fe}=a\left(mol\right);n_{Zn}=b\left(mol\right)\)
\(\Rightarrow56a+65b=12,1\)
Từ (1);(2)\(\Rightarrow\Sigma_{n_{H_2}}=a+b=\dfrac{4,48}{22,4}=0,2\)
Ta có hệ: \(\left\{{}\begin{matrix}56a+65b=12,1\\a+b=0,2\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}a=0,1\\b=0,1\end{matrix}\right.\)
\(\%m_{Fe}=\dfrac{0,1.56}{12,1}.100\%=46,28\%\)
\(\%m_{Zn}=\dfrac{0,1.65}{12,1}.100\%=53,72\%\)
b) Từ (1) và (2) \(\Rightarrow\Sigma n_{H_2SO_4}=a+b=0,2\left(mol\right)\)
\(C_{M\left(H_2SO_4\right)}=\dfrac{0,2}{0,5}=0,4\left(M\right)\).