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a) Gọi \(\left\{{}\begin{matrix}n_{Fe}=a\left(mol\right)\\n_{Zn}=b\left(mol\right)\end{matrix}\right.\) \(\Rightarrow56a+65b=12,1\) (1)
Ta có: \(n_{H_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\)
Bảo toàn electron: \(2n_{Fe}+2n_{Zn}=2n_{H_2}\) \(\Rightarrow2a+2b=0,4\) (2)
Từ (1) và (2) \(\Rightarrow\left\{{}\begin{matrix}a=0,1\\b=0,1\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}\%m_{Fe}=\dfrac{0,1\cdot56}{12,1}\cdot100\%\approx46,28\%\\\%m_{Zn}=53,72\%\end{matrix}\right.\)
b)
Bảo toàn nguyên tố: \(\left\{{}\begin{matrix}n_{FeSO_4}=n_{Fe}=n_{Zn}=n_{ZnSO_4}=0,1\left(mol\right)\\n_{H_2SO_4\left(p.ứ\right)}=n_{H_2}=0,2\left(mol\right)\Rightarrow\Sigma n_{H_2SO_4}=0,2\cdot110\%=0,22\left(mol\right)\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}m_{FeSO_4}=0,1\cdot152=15,2\left(g\right)\\m_{ZnSO_4}=0,1\cdot161=16,1\left(g\right)\\m_{H_2}=0,2\cdot2=0,4\left(g\right)\\m_{H_2SO_4\left(dư\right)}=\left(0,22-0,2\right)\cdot98=1,96\left(g\right)\end{matrix}\right.\)
Mặt khác: \(m_{dd}=m_{KL}+m_{ddH_2SO_4}-m_{H_2}=211,7\left(g\right)\)
\(\Rightarrow\left\{{}\begin{matrix}C\%_{FeSO_4}=\dfrac{15,2}{211,7}\cdot100\%\approx7,18\%\\C\%_{ZnSO_4}=\dfrac{16,1}{211,7}\cdot100\%\approx7,61\%\\C\%_{H_2SO_4}=\dfrac{1,96}{22,4}\cdot100\%\approx0,93\%\end{matrix}\right.\)
\(n_{H_2}=\dfrac{3,36}{22,4}=0,15mol\)
\(\left\{{}\begin{matrix}n_{Al}=x\left(mol\right)\\n_{Ag}=y\left(mol\right)\end{matrix}\right.\Rightarrow27x+108y=4,2\left(1\right)\)
\(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\)
0,1 0,05 0,15
\(\Rightarrow m_{Al}=0,1\cdot27=2,7g\)
\(\Rightarrow m_{Ag}=4,2-2,7=1,5g\)
a)\(\%m_{Al}=\dfrac{2,7}{4,2}\cdot100\%=64,28\%\)
\(\%m_{Ag}=100\%-64,28\%=35,72\%\)
b)\(m_{muối}=0,05\cdot342=17,1g\)
a) Sửa đề: dd H2SO4 9,8%
Ta có: \(n_{H_2}=\dfrac{7,84}{22,4}=0,35\left(mol\right)\) \(\Rightarrow m_{H_2}=0,35\cdot2=0,7\left(g\right)\)
Bảo toàn nguyên tố: \(n_{H_2SO_4}=n_{H_2}=0,35\left(mol\right)\) \(\Rightarrow m_{ddH_2SO_4}=\dfrac{0,35\cdot98}{9,8\%}=350\left(g\right)\)
\(\Rightarrow m_{dd}=m_{KL}+m_{H_2SO_4}-m_{H_2}=361,6\left(g\right)\)
b) Tương tự câu a
a) PTHH: \(Mg+2HCl\rightarrow MgCl_2+H_2\uparrow\) (1)
\(Zn+2HCl\rightarrow ZnCl_2+H_2\uparrow\) (2)
b) Ta có: \(\Sigma n_{H_2}=\dfrac{8,96}{22,4}=0,4\left(mol\right)\)
Gọi số mol của Mg là \(a\) \(\Rightarrow n_{H_2\left(1\right)}=a\)
Gọi số mol của Zn là b \(\Rightarrow n_{H_2\left(2\right)}=b\)
Ta lập được hệ phương trình:
\(\left\{{}\begin{matrix}a+b=0,4\\24a+65b=17,8\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}a=0,2\\b=0,2\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}\%m_{Mg}=\dfrac{0,2\cdot24}{17,8}\cdot100\%\approx26,97\%\\\%m_{Zn}\approx73,03\%\end{matrix}\right.\)
Chất không tan là Ag.
=> mAg= 6,25(g)
nH2=0,25(mol)
PTHH: Zn + H2SO4 -> ZnSO4 + H2
-> nZn=nH2= 0,25(mol)
=>mZn= 0,25 . 65=16,25(g)
=>
%mAg = \(\dfrac{6,25}{6 , 25 + 16 , 25}\) . 100 ≈ 27,778%
⇒% mZn ≈ 72,222%
\(a,PTHH:Zn+2HCl\to ZnCl_2+H_2\\ \Rightarrow n_{Zn}=n_{H_2}=\dfrac{3,7185}{24,79}=0.,15(mol)\\ \Rightarrow m_{Zn}=0,15.65=9,75(g)\\ \Rightarrow \%_{Zn}=\dfrac{9,75}{10}.100\%=97,5\%\\ \Rightarrow \%_{Cu}=100\%-97,5\%=2,5\%\\ b,n_{HCl}=2n_{H_2}=0,3(mol)\\ \Rightarrow m_{dd_{HCl}}=\dfrac{0,3.36,5}{14\%}=78,21(g)\)
\(n_{H2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\)
a) Pt : \(Zn+2HCl\rightarrow ZnCl_2+H_2|\)
1 2 1 1
0,2 0,2
b) \(n_{Zn}=\dfrac{0,2.1}{1}=0,2\left(mol\right)\)
\(m_{Zn}=0,2.65=13\left(g\right)\)
\(m_{Cu}=19,4-13=6,4\left(g\right)\)
Chúc bạn học tốt
\(a)CTHH:ZnSO_4\\ b)n_{Zn}=\dfrac{0,65}{65}=0,01mol\\ Zn+H_2SO_4\rightarrow ZnSO_4+H_2\)
0,01 0,01
\(V_{H_2\left(đktc\right)}=0,01.22,4=0,224l\\ V_{H_2\left(đkc\right)}=0,01.24,79=0,2479l\)
\(n_{Zn}=\dfrac{0,65}{65}=0,01\left(mol\right)\\ a,Zn+H_2SO_4\rightarrow ZnSO_4+H_2\\ b,n_{H_2}=n_{Zn}=0,01\left(mol\right)\\ Vậy:V_{H_2\left(đktc\right)}=0,01.22,4=0,224\left(l\right)\\ V_{H_2\left(đkc\right)}=0,01.24,79=0,2479\left(l\right)\)