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\(a.n_{H_2}=\dfrac{5,6}{22,4}=0,25\left(mol\right)\\ Đặt:\left\{{}\begin{matrix}n_{Al}=a\left(mol\right)\\n_{Mg}=b\left(mol\right)\end{matrix}\right.\left(a,b>0\right)\\ 2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\\ Mg+H_2SO_4\rightarrow MgSO_4+H_2\\ \rightarrow\left\{{}\begin{matrix}27a+24b=5,1\\1,5a+b=0,25\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=0,1\\b=0,1\end{matrix}\right.\\ \left\{{}\begin{matrix}\%m_{Al}=\dfrac{27.0,1}{5,1}.100\approx52,941\%\\\%m_{Mg}\approx47,059\%\end{matrix}\right.\)
\(b.m_{ddH_2SO_4}=\dfrac{0,25.98.100}{9,8}=250\left(g\right)\\ m_{ddsau}=m_{Al,Mg}+m_{ddH_2SO_4}-m_{H_2}=5,1+250-0,25.2=254,6\left(g\right)\\ C\%_{ddAl_2\left(SO_4\right)_3}=\dfrac{0,05.342}{254,6}.100\approx6,716\%\\ C\%_{ddMgSO_4}=\dfrac{0,1.120}{254,6}.100\approx4,713\%\)
Gọi nMg = x
nAl = y (mol)
\(\left\{{}\begin{matrix}24x+27y=5,1\\x+1,5y=0,25\end{matrix}\right.\)
\(\rightarrow x=0,1;y=0,1\)
\(\%m_{Mg}=\dfrac{0,1.24}{5,1}.100\%\approx47,06\%\)
\(\%m_{Al}=100\%-47,06\%=52,94\%\)
\(m_{H_2SO_4}=\dfrac{\left(0,1+0,15\right).98.100}{10}=245\left(g\right)\)
\(n_{H_2}=\dfrac{8,96}{22,4}=0,4(mol)\\ n_{Al}=x(mol);n_{Mg}=y(mol)\\ \Rightarrow 27x+24y=7,8(1)\\ a,2Al+6HCl\to 2AlCl_3+3H_2\\ Mg+2HCl\to MgCl_2+H_2\\ \Rightarrow 1,5x+y=0,4(2)\\ (1)(2)\Rightarrow x=0,2(mol);y=0,1(mol)\\ \Rightarrow \%_{Al}=\dfrac{0,2.27}{7,8}.100\%=69,23\%\\ \Rightarrow \%_{Mg}=100\%-69,23\%=30,77\%\)
\(b,\Sigma n_{HCl}=3x+2y=0,8(mol)\\ \Rightarrow C\%_{HCl}=\dfrac{0,8.36,5}{192,2}.100\%=15,19\%\\ c,n_{AlCl_3}=0,2(mol);n_{MgCl_2}=0,1(mol)\\ \Rightarrow C\%_{AlCl_3}=\dfrac{0,2.133,5}{0,2.27+192,2-0,3.2}.100\%=13,55\%\\ C\%_{MgCl_2}=\dfrac{0,1.95}{0,1.24+192,2-0,1.2}.100\%=4,89\%\)
nH2nH2=1,34422,41,34422,4=0,06 (mol)
Cu+HCl→Cu+HCl→ ko pứ
Fe+2HCl→FeCl2+H2↑Fe+2HCl→FeCl2+H2↑
0,06 0,12 ←0,06 (mol)
%mFe=0,06.56/13.100 % ≈25,85 %
%mCu=100 % - 25,85 %=74,15 %
///
mctHCll=0,12.26,5=4,38 (g)
mddHCll=4,38.100/15 =29,2 (g)
\(Fe+2HCl\rightarrow FeCl_2+H_2\)
\(Zn+2HCl\rightarrow ZnCl_2+H_2\)
Gọi \(n_{Fe}=x\left(mol\right);n_{Zn}=y\left(mol\right)\)
\(Tacó:\left\{{}\begin{matrix}127x+136y=10,43\\2x+2y=0,5.0,4=0,2\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}x=0,35\\y=-0,25\end{matrix}\right.\)
Xem lại đề !
em có thể tiếp tục sân chơi được không? Cố lên ~~~ anh đi làm cv đây
\(n_{Mg}=x(mol);n_{Fe}=y(mol)\\ \Rightarrow 24x+56y=4(1)\\ n_{H_2}=\dfrac{2,24}{22,4}=0,1(mol)\\ Mg+2HCl\to MgCl_2+H_2\\ Fe+2HCl\to FeCl_2+H_2\\ \Rightarrow x+y=0,1(2)\\ (1)(2)\Rightarrow x=y=0,05(mol)\\ \Rightarrow \%_{Fe}=\dfrac{0,05.56}{4}.100\%=70\%\\ \Rightarrow \%_{Mg}=100\%-70\%=30\%\)
Gọi $n_{Mg} = n_{Al} = a(mol) ; n_{Fe} = b(mol)$
Ta có :
$24a + 27a + 56b = 15,8(1)$
$n_{HCl} > 2n_{H_2}$ nên HCl dư
Ta có :
$n_{H_2} = a + 1,5a + b = \dfrac{13,44}{22,4} = 0,6(2)$
Từ (1)(2) suy ra a = 0,2 ; b = 0,1
$\%m_{Al} = \dfrac{0,2.27}{15,8}.100\% = 34,18\%$
$\%m_{Mg} = \dfrac{0,2.24}{15,8}.100\% = 30,38\%$
$\%m_{Fe} = 35,44\%$
$n_{HCl\ pư} = 2n_{H_2} = 1,2(mol)$
Bảo toàn khối lượng : $m_{muối} = 15,8 + 1,2.36,5 - 0,6.2 = 58,4(gam)$
a) \(n_{Al}=\dfrac{5,4}{27}=0,2\left(mol\right)\)
PTHH: 2Al + 3H2SO4 --> Al2(SO4)3 + 3H2
______0,2---->0,3------------>0,1------>0,3______(mol)
=> VH2 = 0,3.22,4= 6,72(l)
b) \(C_{M\left(H_2SO_4\right)}=\dfrac{0,3}{0,1}=3M\)
\(C_{M\left(Al_2\left(SO_4\right)_3\right)}=\dfrac{0,1}{0,1}=1M\)
Câu 3:
\(n_{H_2}=\dfrac{6,72}{22,4}=0,3(mol)\\ PTHH:Mg+2HCl\to MgCl_2+H_2\\ MgO+2HCl\to MgCl_2+H_2O\\ \Rightarrow n_{Mg}=n_{H_2}=0,3(mol)\\ \Rightarrow \%_{Mg}=\dfrac{0,3.24}{15,2}.100\%=47,37\%\\ \Rightarrow \%_{MgO}=100\%-47,37\%=52,63\%\)
\(n_{MgO}=\dfrac{15,2-0,3.24}{40}=0,2(mol)\\ \Rightarrow \Sigma n_{HCl}=0,3.2+0,2.2=1(mol)\\ \Rightarrow m_{dd_{HCl}}=\dfrac{1.36,5}{10\%}=365(g)\\ \Sigma n_{MgCl_2}=0,2+0,3=0,5(mol)\\ \Rightarrow C\%_{MgCl_2}=\dfrac{0,5.95}{15,2+365}.100\%=12,49\%\)
\(PTHH:Mg+2H_2SO_{4(đ)}\to MgSO_4+2H_2O+SO_2\uparrow\\ MgO+H_2SO_4\to MgSO_4+H_2O\\ \Rightarrow n_{SO_2}=n_{Mg}=0,3(mol)\\ \Rightarrow V_{SO_2}=0,3.22,4=6,72(l)\)
Mg +H2SO4--->MgSO4 +H2
x x x x mol
Fe+ H2SO4---> FeSO4+ H2
y y y y mol
theo bài ta có : 24x+ 56y=1,36 và x+y=0,672/22,4
=> x=0,01 mol và y=0,02 mol
=> mMg=0,24 gam mFe=1,12 gam
tớ thấy đề bài khó để là ý b) bạn ạ nếu bạn xem lạ đề bài thì tốt quá