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\(n_{H_2}=\dfrac{8,96}{22,4}=0,4(mol)\\ n_{Al}=x(mol);n_{Mg}=y(mol)\\ \Rightarrow 27x+24y=7,8(1)\\ a,2Al+6HCl\to 2AlCl_3+3H_2\\ Mg+2HCl\to MgCl_2+H_2\\ \Rightarrow 1,5x+y=0,4(2)\\ (1)(2)\Rightarrow x=0,2(mol);y=0,1(mol)\\ \Rightarrow \%_{Al}=\dfrac{0,2.27}{7,8}.100\%=69,23\%\\ \Rightarrow \%_{Mg}=100\%-69,23\%=30,77\%\)
\(b,\Sigma n_{HCl}=3x+2y=0,8(mol)\\ \Rightarrow C\%_{HCl}=\dfrac{0,8.36,5}{192,2}.100\%=15,19\%\\ c,n_{AlCl_3}=0,2(mol);n_{MgCl_2}=0,1(mol)\\ \Rightarrow C\%_{AlCl_3}=\dfrac{0,2.133,5}{0,2.27+192,2-0,3.2}.100\%=13,55\%\\ C\%_{MgCl_2}=\dfrac{0,1.95}{0,1.24+192,2-0,1.2}.100\%=4,89\%\)
PT: \(Mg+H_2SO_4\rightarrow MgSO_4+H_2\)
\(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\)
a, Giả sử: \(\left\{{}\begin{matrix}n_{Mg}=x\left(mol\right)\\n_{Al}=y\left(mol\right)\end{matrix}\right.\)
⇒ 24x + 27y = 12,6 (1)
Ta có: \(n_{H_2}=\dfrac{13,44}{22,4}=0,6\left(mol\right)\)
Theo PT: \(n_{H_2}=n_{Mg}+\dfrac{3}{2}n_{Al}=x+\dfrac{3}{2}y\left(mol\right)\)
\(\Rightarrow x+\dfrac{3}{2}y=0,6\left(2\right)\)
Từ (1) và (2) \(\Rightarrow\left\{{}\begin{matrix}x=0,3\left(mol\right)\\y=0,2\left(mol\right)\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\%m_{MG}=\dfrac{0,3.24}{12,6}.100\%\approx57,1\%\\\%m_{Al}\approx42,9\%\end{matrix}\right.\)
b, Theo PT: \(\left\{{}\begin{matrix}n_{H_2SO_4}=n_{H_2}=0,6\left(mol\right)\\n_{MgSO_4}=n_{Mg}=0,3\left(mol\right)\\n_{Al_2\left(SO_4\right)_3}=\dfrac{1}{2}n_{Al}=0,1\left(mol\right)\end{matrix}\right.\)
\(\Rightarrow m_{H_2SO_4}=0,6.98=58,8\left(g\right)\Rightarrow m_{ddH_2SO_4}=\dfrac{58,8}{14,7\%}=400\left(g\right)\)
Ta có: m dd sau pư = 12,6 + 400 - 0,6.2 = 411,4 (g)
\(\Rightarrow\left\{{}\begin{matrix}C\%_{MgSO_4}=\dfrac{0,3.120}{411,4}.100\%\approx8,75\%\\C\%_{Al_2\left(SO_4\right)_3}=\dfrac{0,1.342}{411,4}.100\%\approx8,31\%\end{matrix}\right.\)
Bạn tham khảo nhé!
Mg+2HCl->MgCl2+H2
a..............................a(mol)
Fe+2HCl->FeCl2+H2
b............................b(mol)
=>nCu=3,2/64=0,05mol
=>%mCu=(3,2.100%)/11,2=28,6%
\(=>\left\{{}\begin{matrix}24a+56b=11,2-3,2\\a+b=0,2\end{matrix}\right.=>\left\{{}\begin{matrix}a=0,1\\b=0,1\end{matrix}\right.\)
=>mMg=24.0,1=2,4g=>%Mg=(2,4.100%)/11,2=21,4%
=>%Fe=100%-21,4%-28,6%=50%
b, MgCl2+2NaOH->Mg(OH)2+2NaCL
FeCl2+2NaOH->Fe(OH)2+2NaCl
=>m(kết tủa)=mMg(OH)2+mFe(OH)2
=0,1(58+90)=14,8g
a) mCu= m(k tan)= 3,2(g)
=> m(Mg, Fe)= 11,2- 3,2=8(g)
nH2= 4,48/22,4=0,2(mol)
PTHH: Mg + 2 HCl -> MgCl2 + H2
a______________2a__a______a(mol)
Fe + 2 HCl -> FeCl2 + H2
b____2b_____b_____b(mol)
Ta có hpt:
\(\left\{{}\begin{matrix}24a+56b=8\\a+b=0,2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=0,1\\b=0,1\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}m_{Mg}=24.0,1=2,4\left(g\right)\\m_{Fe}=56.0,1=5,6\left(g\right)\end{matrix}\right.\)
=> %mMg= (2,4/11,2).100=21,429%
%mFe= (5,6/11,2).100=50%
=>%mCu= (3,2/11,2).100=28,571%
b/ MgCl2 + 2 NaOH -> Mg(OH)2 + 2 NaCl
0,1___________________0,1(mol)
FeCl2 + 2 NaOH -> Fe(OH)2 +2 NaCl
0,1__________________0,1(mol)
m(kt)=mMg(OH)2 + mFe(OH)2= 58.0,1+ 90.0,1= 14,8(g)
PTHH:
\(Mg+H_2SO_4\rightarrow MgSO_4+H_2\)
\(Zn+H_2SO_4\rightarrow ZnSO_4+H_2\)
Thí nghiệm 1:
Ta có:
\(n_{H_2SO_4}=n_{H_2}=\dfrac{8,96}{22,4}=0,4\left(mol\right)\)
\(\Rightarrow C_{M\left(H_2SO_4\right)}=\dfrac{0,4}{2}=0,2M\)
\(n_{Mg}+n_{Zn}=n_{H_2SO_4}=0,4\left(mol\right)\left(1\right)\)
Lại có: \(24n_{Mg}+65n_{Zn}=24,3\left(2\right)\)
Giải hệ hai phương trình (1) và (2) ta được:
\(\left\{{}\begin{matrix}n_{Mg}=0,04\left(mol\right)\\n_{Zn}=0,36\left(mol\right)\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}m_{Mg}=0,96\left(g\right)\\n_{Zn}=23,4\left(g\right)\end{matrix}\right.\)
Thí nghiệm 2:
Ta có:
\(n_{H_2SO_4}=n_{H_2}=\dfrac{11,2}{22,4}=0,5\left(mol\right)\)
\(\Rightarrow C_{M\left(H_2SO_4\right)}=\dfrac{0,5}{2}=0,25M\)
\(n_{Mg}+n_{Zn}=n_{H_2SO_4}=0,5\left(mol\right)\left(3\right)\)
Lại có: \(24n_{Mg}+65n_{Zn}=24,3\left(4\right)\)
Giải hệ hai phương trình (3) và (4) ta được:
\(\left\{{}\begin{matrix}n_{Mg}=0,2\left(mol\right)\\n_{Zn}=0,3\left(mol\right)\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}m_{Mg}=4,8\left(g\right)\\n_{Zn}=19,5\left(g\right)\end{matrix}\right.\)
vì sao CM và khối lượng lại ra 2 kết quả vậy ạ?