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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 )
\(\left\{{}\begin{matrix}C_{M_{Al_2\left(SO_4\right)_3}}=\dfrac{0,075}{0,8}=0,09M\\C_{M_{ZnSO_4}}=\dfrac{0,1}{0,8}=0,125M\end{matrix}\right.\)
\(n_{SO_2}=\dfrac{V_{SO_2}}{22,4}=\dfrac{2,24}{22,4}=0,1mol\)
Gọi \(\left\{{}\begin{matrix}n_{Mg}=x\\n_{Cu}=y\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}m_{Mg}=24x\\m_{Cu}=64y\end{matrix}\right.\)
\(Mg+2H_2SO_4\rightarrow MgSO_4+SO_2+2H_2O\)
x 2x x ( mol )
\(Cu+2H_2SO_4\rightarrow CuSO_4+SO_2+2H_2O\)
y 2x y ( mol )
Ta có:
\(\left\{{}\begin{matrix}24x+64y=4,4\\x+y=0,1\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=0,05\\y=0,05\end{matrix}\right.\)
\(\Rightarrow m_{Mg}=0,05.24=1,2g\)
\(\Rightarrow m_{Cu}=0,05.64=3,2\)
\(m_{H_2SO_4}=n_{H_2SO_4}.M_{H_2SO_4}=\left(2.0,05+2.0,05\right).98=0,2.98=19,6g\)
À thêm đk H2SO4 đặc nóng nhá chứ H2SO4 loãng thì PTHH là:
Mg + H2SO4 -> MgSO4 + H2
Cu không tác dụng với H2SO4 loãng
Ta có: \(m_{H_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\) \(\Rightarrow m_{H_2}=0,2\cdot2=0,4\left(g\right)\)
\(\Rightarrow m_{dd\left(sau.pư\right)}=m_{hh}+m_{ddH_2SO_4}-m_{H_2}=309,6\left(g\right)\)
\(\Rightarrow a=309,6-300=9,6\left(g\right)\)
Giả sử: \(\left\{{}\begin{matrix}n_{Mg}=a\left(mol\right)\\n_{Zn}=b\left(mol\right)\\n_{Cu}=c\left(mol\right)\\n_{Al}=d\left(mol\right)\end{matrix}\right.\)
_ Khi tác dụng với HCl.
Ta có: \(n_{H_2}=\dfrac{10,08}{22,4}=0,45\left(mol\right)\)
Theo ĐLBT mol e, có: 2a + 2b + 3d = 0,45.2 ⇒ 2a + 2b + 3d = 0,9 (1)
_ Khi tác dụng với H2SO4 đặc nóng.
Ta có: \(n_{SO_2}=\dfrac{10,64}{22,4}=0,475\left(mol\right)\)
Theo ĐLBT mol e, có: 2a + 2b + 2c + 3d = 0,475.2
⇒ 2a + 2b - 2c + 3d = 0,95 (2)
Trừ 2 vế của (1) và (2), có: c = 0,025 (mol)
\(\Rightarrow\%m_{Cu}=\dfrac{0,025.64}{14,7}.100\%\approx10,88\%\)
Bạn tham khảo nhé!
Đặt \(\left\{{}\begin{matrix}n_{Mg}=a\left(mol\right)\\n_{Zn}=b\left(mol\right)\\n_{Cu}=c\left(mol\right)\\n_{Al}=d\left(mol\right)\end{matrix}\right.\)
Ta có: \(\left\{{}\begin{matrix}n_{H_2}=\dfrac{10,08}{22,4}=0,45\left(mol\right)\\n_{SO_2}=\dfrac{10,64}{22,4}=0,475\left(mol\right)\end{matrix}\right.\)
Bảo toàn electron: \(\left\{{}\begin{matrix}2a+2b+2c+3d=0,475\cdot2\\2a+2b+3d=0,45\cdot2\end{matrix}\right.\)
\(\Rightarrow2c=0,475\cdot2-0,45\cdot2=0,05\) \(\Rightarrow c=0,025\)
\(\Rightarrow\%m_{Cu}=\dfrac{0,025\cdot64}{14,7}\cdot100\%\approx10,88\%\)
