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Ta có: mH\(_2\)SO\(_4\)= 200 . 14% = 28g
=> nH\(_2\)SO\(_4\)= 28/98 = 0,285 (mol)
PTHH : 3H\(_2\)SO\(_4\) + 2 Al ----> Al\(_2\)(SO\(_4\))\(_3\) + 3H\(_2\)
n H\(_2\)SO\(_4\)= 3nAl\(_2\)(SO\(_4\))\(_3\)
=> nAl\(_2\)(SO\(_4\))\(_3\)= 0,285 : 3 =0,095 (mol)
=> mAl\(_2\)(SO\(_4\))\(_3\)= 0,095 . 342 = 32,49 g
\(n_{Al_2\left(SO_4\right)_3}=\dfrac{3,42}{342}=0,01mol\)
\(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\)
0,03 0,01 ( mol )
\(V_{H_2SO_4}=\left(\dfrac{0,03}{0,04}\right)+0,1=0,85l\)
\(n_{H_2}=\dfrac{2,464}{22,4}=0,11mol\)
\(\left\{{}\begin{matrix}Al:x\left(mol\right)\\Fe:y\left(mol\right)\end{matrix}\right.\Rightarrow Muối\left\{{}\begin{matrix}Al_2\left(SO_4\right)_3\\FeSO_4\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}BTe:3x+2y=2n_{H_2}=0,22\\\dfrac{x}{2}\cdot342+y\cdot152=14,44\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0,04mol\\y=0,05mol\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}m_{Al}=0,04\cdot27=1,08g\\m_{Fe}=0,05\cdot56=2,8g\end{matrix}\right.\)
\(Al_2\left(SO_4\right)_3+3BaCl_2\rightarrow2AlCl_3+3BaSO_4\downarrow\)
0,02 0,06
\(FeSO_4+BaCl_2\rightarrow BaSO_4\downarrow+FeCl_2\)
0,05 0,05
\(\Rightarrow\Sigma n_{\downarrow}=0,06+0,05=0,11\Rightarrow m_{BaSO_4}=x=25,63g\)
nH2 =\(\frac{0.561}{22.4}\)\(\approx\) 0.025 (mol)
Đặt số mol của Al là x (mol)
số mol của Fe là y (mol) (x,y > 0)
Al0 ----------> Al+3 + 3e 2H+1 +2e ----------> H20
x ----------------> 3x 0.05 0.02 (mol)
Fe0 ---------> Fe+2 + 2e
y -----------------> 2y
Ta có: \(\Sigma n_{e_{ }cho}=\Sigma n_{e_{ }nh\text{ận}}\)
<=> 3x + 2y = 0.05 (*)
Theo giả thiết và (*) ta có hệ: \(\left\{\begin{matrix}3x+2y=0.05\\27x+56y=0.83\end{matrix}\right.\)
<=> \(\left\{\begin{matrix}x=0.01\\y=0.01\end{matrix}\right.\)
=> %mAl = \(\frac{0.01.27.100\%}{0.83}\)\(\approx\) 32.53%
%mFe = 100% - 32.53% \(\approx\) 67.47%
\(n_{H_2}=\dfrac{8,96}{22,4}=0,4\left(mol\right)\\ PTHH:2Al+6HCl\rightarrow2AlCl_3+3H_2\\ Fe+2HCl\rightarrow FeCl_2+3H_2\\ Đặt:n_{Al}=a\left(mol\right);n_{Fe}=b\left(mol\right)\left(a,b>0\right)\\ \Rightarrow\left\{{}\begin{matrix}27a+56b=11\\1,5a+b=0,4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=0,2\\b=0,1\end{matrix}\right.\\ a,n_{HCl}=2.n_{H_2}=2.0,4=0,8\left(mol\right)\\ \Rightarrow V_{ddHCl}=\dfrac{0,8}{8}=0,1\left(l\right)\\ b,FeCl_2+2AgNO_3\rightarrow Fe\left(NO_3\right)_2+2AgCl\downarrow\\ AlCl_3+3AgNO_3\rightarrow Al\left(NO_3\right)_3+3AgCl\downarrow\\ n_{AgCl}=n_{AgNO_3}=3.n_{AlCl_3}+2.n_{FeCl_2}=3.a+2.b=3.0,2+2.0,1=0,8\left(mol\right)\\ \Rightarrow a=\dfrac{170.0,8}{250}.100=54,4\%\\ b=m_{\downarrow}=m_{AgCl}=0,8.143,5=114,8\left(g\right)\)
Ta có : \(m_{H_2SO_4}=14\%.200=28\left(g\right)\)
\(\Rightarrow n_{H_2SO_4}=\frac{28}{98}=\frac{2}{7}\left(mol\right)\)
PTHH : \(2Al+3H_2SO_4=Al_2\left(SO_4\right)_3+3H_2\uparrow\)
(mol) 2/7 2/21
\(\Rightarrow m_{Al_2\left(SO_4\right)_3}=\frac{2}{21}.342=\frac{228}{7}\left(g\right)\)