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$a)PTHH:2Al+6HCl\to 2AlCl_3+3H_2$
$n_{H_2}=\dfrac{5,04}{22,4}=0,225(mol)$
$\Rightarrow n_{Al}=0,15(mol)$
$\Rightarrow \%m_{Al}=\dfrac{0,15.27}{9,45}.100\%\approx 42,86\%$
$\Rightarrow \%m_{Cu}=100-42,86=57,14\%$
$b)$ Theo PT: $n_{HCl}=2n_{H_2}=0,45(mol)$
$\Rightarrow C_{M_{HCl}}=\dfrac{0,45.110\%}{0,5}=0,99M$
PTHH: \(2Fe+6H_2SO_{4\left(đ\right)}\underrightarrow{t^o}Fe_2\left(SO_4\right)_3+3SO_2\uparrow+6H_2O\)
\(CuO+H_2SO_4\rightarrow CuSO_4+H_2O\)
a) Ta có: \(n_{SO_2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\) \(\Rightarrow n_{Fe}=\dfrac{1}{15}\left(mol\right)\)
\(\Rightarrow\%m_{Fe}=\dfrac{\dfrac{1}{15}\cdot56}{13,6}\cdot100\%\approx27,45\%\) \(\Rightarrow\%m_{CuO}=72,55\%\)
b) Ta có: \(m_{CuO}=13,6-\dfrac{1}{15}\cdot56\approx9,9\left(g\right)\) \(\Rightarrow n_{CuO}=n_{H_2SO_4}=\dfrac{9,9}{80}=0,12375\left(mol\right)\)
*Làm gì có H2SO4 loãng đâu nhỉ ??
a.\(n_{H_2}=\dfrac{7,28}{22,4}=0,325mol\)
Gọi \(\left\{{}\begin{matrix}n_{Al}=x\\n_{Zn}=y\end{matrix}\right.\) \(\left(mol\right)\) \(\rightarrow27x+65y=10,55\left(g\right)\) (1)
\(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\)
x 1/2 x 3/2 x ( mol )
\(Zn+H_2SO_4\rightarrow ZnSO_4+H_2\)
y y y ( mol )
\(\rightarrow\dfrac{3}{2}x+y=0,325\left(mol\right)\) (2)
\(\left(1\right);\left(2\right)\rightarrow\left\{{}\begin{matrix}x=0,15\\y=0,1\end{matrix}\right.\)
\(\rightarrow\left\{{}\begin{matrix}\%m_{Al}=\dfrac{0,15.27}{10,55}.100\%=38,38\%\\\%m_{Zn}=100\%-38,38\%=61,62\%\end{matrix}\right.\)
b.\(\left\{{}\begin{matrix}n_{Al_2\left(SO_4\right)_3}=\dfrac{1}{2}.0,15=0,075\\n_{ZnSO_4}=0,1\end{matrix}\right.\) ( mol )
\(\left\{{}\begin{matrix}C_{M_{Al_2\left(SO_4\right)_3}}=\dfrac{0,075}{0,8}=0,09M\\C_{M_{ZnSO_4}}=\dfrac{0,1}{0,8}=0,125M\end{matrix}\right.\)
nHCl = 0,3.0,3 = 0,09 (mol)
\(n_{H_2}=\dfrac{0,672}{22,4}=0,03\left(mol\right)\)
PTHH: 2Al + 6HCl --> 2AlCl3 + 3H2
0,02<-0,06<------------0,03
CuO + 2HCl --> CuCl2 + H2O
0,015<-0,03
=> \(\left\{{}\begin{matrix}m_{Al}=0,02.27=0,54\left(mol\right)\\m_{CuO}=0,015.80=1,2\left(g\right)\end{matrix}\right.\)
\(n_{HCl}=0,3\cdot0,3=0,09mol\)
\(n_{H_2}=\dfrac{0,672}{22,4}=0,03mol\)
\(CuO+2HCl\rightarrow CuCl_2+H_2O\)
\(2Al+6HCl\rightarrow2AlCl_3+3H_2\)
0,02 0,06 0,03
\(\Rightarrow n_{HCl\left(CuO\right)}=0,09-0,06=0,03mol\)
\(\Rightarrow n_{CuO}=n_{HCl}=0,03mol\) (theo pt)
\(\Rightarrow m_{CuO}=0,03\cdot80=2,4g\)
\(m_{Al}=0,02\cdot27=0,54g\)
\(n_{H_2}=\dfrac{0,672}{22,4}=0,03\left(mol\right)\)
PTHH:
2Al + 6HCl ---> 2AlCl3 + 3H2
0,02 0,06 0,03
nHCl = 0,3.0,3 = 0,09 (mol)
nHCl (CuO) = 0,09 - 0,06 = 0,03 (mol)
CuO + 2HCl ---> CuCl2 + H2O
0,015 0,03
\(\rightarrow\left\{{}\begin{matrix}m_{Al}=0,02.27=0,54\left(g\right)\\m_{CuO}=0,015.80=1,2\left(g\right)\end{matrix}\right.\)
P/s: mình có thấy chị Hương Giang làm nhưng sai phần tính số mol của CuO "\(n_{CuO}=n_{HCl}\) (theo pt)"
Đổi: 400ml = 0,4l
nHCl = CM.V = 0,4 (mol)
Pthh:
FeO + 2HCl -> FeCl2 + H2O
0,025 0,05 0,025
CaCO3 + 2HCl -> CaCl2 + H2O + CO2
0,1 0,2 0,1 0,1
nCO2 = V/22,4 = 2,24/22,4 = 0,1 (mol)
=> mCaCO3 = M.n = 100 x 0,1 = 10 (g)
=> mFeO = 1,8 (g) => nFeO = 0,025 (mol)
=> nHCl(dư) = 0,4 - 0,2 - 0,05 = 0,15 (mol)
+) CMHCl(dư) = n/V = 0,15/0,4 = 0,375 mol
+) CMFeCl2 = n/V = 0,025/4 = 0,0625 mol
+) CMCaCl2 = n/V = 0,1/4 = 0,25 mol
a) Ta có: \(\left\{{}\begin{matrix}n_{Cu}=\dfrac{0,32}{64}=0,005\left(mol\right)\\n_{H_2}=\dfrac{0,448}{22,4}=0,02\left(mol\right)\end{matrix}\right.\)
Đặt \(\left\{{}\begin{matrix}n_{Fe}=a\left(mol\right)\\n_{Al}=b\left(mol\right)\end{matrix}\right.\) \(\Rightarrow56a+27b=0,87-0,32=0,55\) (1)
Bảo toàn electron: \(2a+3b=2n_{H_2}=0,04\) (2)
Từ (1) và (2) \(\Rightarrow\left\{{}\begin{matrix}a=0,005\\b=0,01\end{matrix}\right.\)
Bảo toàn nguyên tố: \(\left\{{}\begin{matrix}n_{FeSO_4}=n_{Fe}=0,005\left(mol\right)\\n_{Al_2\left(SO_4\right)_3}=\dfrac{1}{2}n_{Al}=0,005\left(mol\right)\end{matrix}\right.\)
\(\Rightarrow C_{M_{FeSO_4}}=\dfrac{0,005}{0,3}\approx0,02\left(M\right)=C_{M_{Al_2\left(SO_4\right)_3}}\)
b)
Ta thấy trong 0,87 gam hh X có 0,005 mol Fe, 0,005 mol Cu và 0,01 mol Al
\(\Rightarrow\) Trong 2,61 gam hh X có 0,015 mol Fe, 0,015 mol Cu và 0,03 mol Al
PTHH: \(2Fe+6H_2SO_{4\left(đặc\right)}\xrightarrow[]{t^o}Fe_2\left(SO_4\right)_3+3SO_2\uparrow+6H_2O\)
\(Cu+2H_2SO_{4\left(đặc\right)}\xrightarrow[]{t^o}CuSO_4+SO_2\uparrow+2H_2O\)
\(2Al+6H_2SO_{4\left(đặc\right)}\xrightarrow[]{t^o}Al_2\left(SO_4\right)_3+3SO_2\uparrow+6H_2O\)
Ta có: \(n_{H_2SO_4}=3n_{Fe}+3n_{Al}+2n_{Cu}=0,165\left(mol\right)\)
\(\Rightarrow m_{ddH_2SO_4}=\dfrac{0,165\cdot98}{78\%}\approx20,73\left(g\right)\)