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Đặt: \(\left\{{}\begin{matrix}x=n_{Fe}\left(mol\right)\\y=n_{Al}\left(mol\right)\end{matrix}\right.\)
\(\sum m_{hh}=11\left(g\right)\Rightarrow56x+27y=11\left(1\right)\)
\(PTHH:Fe+2HCl\rightarrow FeCl_2+H_2\uparrow\\ \left(mol\right)....x\rightarrow..2x........x......x\\ PTHH:2Al+6HCl\rightarrow2AlCl_3+3H_2\uparrow\\ \left(mol\right)....y\rightarrow..3y.........y......1,5y\)
\(n_{H_2}=\dfrac{8,96}{22,4}=0,4\left(mol\right)\\ \Rightarrow x+1,5y=0,4\left(2\right)\)
\(\xrightarrow[\left(2\right)]{\left(1\right)}\left\{{}\begin{matrix}x=0,1\\y=0,2\end{matrix}\right.\)
a) \(\%m_{Fe}=\dfrac{56.0,1}{11}=51\%\)
\(\rightarrow\%m_{Al}=49\%\)
b) \(\sum m_{ctHCl}=\left(2.0,1+3.0,2\right).36,5=29,2\left(g\right)\)
\(m_{ddHCl}=\dfrac{29,2.100\%}{10\%}=292\left(g\right)\)
c) \(m_{H_2\uparrow}=\left(1.0,1+1,5.0,2\right).2=0,8\left(g\right)\)
\(m_{ddsaupu}=m_{hh}+m_{ddHCl}-m_{H_2\uparrow}=11+292-0,8=302,2\left(g\right)\)
\(C\%_{FeCl_2}=\dfrac{0,1.127}{302,2}.100=4,2\%\\ C\%_{AlCl_3}=\dfrac{0,2.133,5}{302,2}.100=8,8\%\)
Đặt: {x=nFe(mol)y=nAl(mol){x=nFe(mol)y=nAl(mol)
∑mhh=11(g)⇒56x+27y=11(1)∑mhh=11(g)⇒56x+27y=11(1)
PTHH:Fe+2HCl→FeCl2+H2↑(mol)....x→..2x........x......xPTHH:2Al+6HCl→2AlCl3+3H2↑(mol)....y→..3y.........y......1,5yPTHH:Fe+2HCl→FeCl2+H2↑(mol)....x→..2x........x......xPTHH:2Al+6HCl→2AlCl3+3H2↑(mol)....y→..3y.........y......1,5y
nH2=8,9622,4=0,4(mol)⇒x+1,5y=0,4(2)nH2=8,9622,4=0,4(mol)⇒x+1,5y=0,4(2)
(1)−→(2){x=0,1y=0,2→(2)(1){x=0,1y=0,2
a) %mFe=56.0,111=51%%mFe=56.0,111=51%
→%mAl=49%→%mAl=49%
b) ∑mctHCl=(2.0,1+3.0,2).36,5=29,2(g)∑mctHCl=(2.0,1+3.0,2).36,5=29,2(g)
mddHCl=29,2.100%10%=292(g)mddHCl=29,2.100%10%=292(g)
c) mH2↑=(1.0,1+1,5.0,2).2=0,8(g)mH2↑=(1.0,1+1,5.0,2).2=0,8(g)
mddsaupu=mhh+mddHCl−mH2↑=11+292−0,8=302,2(g)mddsaupu=mhh+mddHCl−mH2↑=11+292−0,8=302,2(g)
\(Đặt:n_{Fe}=a\left(mol\right);n_{Al}=b\left(mol\right)\left(a,b>0\right)\\ Kim.loại.còn.lại.sau.p.ứ:Cu\\ n_{Cu}=\dfrac{25,4}{64}=0,4\left(mol\right)\\ a,PTHH:Fe+CuSO_4\rightarrow FeSO_4+Cu\\ 2Al+3CuSO_4\rightarrow Al_2\left(SO_4\right)_3+3Cu\\ \Rightarrow\left\{{}\begin{matrix}56a+27b=11\\a+1,5b=0,4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=0,1\\b=0,2\end{matrix}\right.\\b, \%m_{Fe}=\dfrac{0,1.56}{11}.100\approx50,909\%\\ \%m_{Cu}\approx100\%-50,909\%\approx49,091\%\\ c,V_{ddsau}=V_{ddCuSO_4}=0,2\left(l\right)\\ C_{MddFeSO_4}=\dfrac{0,1}{0,2}=0,5\left(M\right);C_{MddAl_2\left(SO_4\right)_3}=\dfrac{0,2:2}{0,2}=0,5\left(M\right)\\ d,C_{MddCuSO_4}=\dfrac{a+1,5b}{0,2}=2\left(M\right)\)
a) \(n_{H_2}=\dfrac{8,96}{22,4}=0,4\left(mol\right)\left(1dm^3=1l\right)\)
PTHH: 2Al + 3H2SO4 → Al2(SO4)3 + 3H2
Mol: x 1,5x
PTHH: Mg + H2SO4 → MgSO4 + H2
Mol: y y
Ta có: \(\left\{{}\begin{matrix}27x+24y=7,8\\1,5x+y=0,4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0,2\\y=0,1\end{matrix}\right.\)
\(\%m_{Al}=\dfrac{0,2.27.100\%}{7,8}=69,23\%;\%m_{Mg}=100-69,23=30,77\%\)
b)
PTHH: 2Al + 3H2SO4 → Al2(SO4)3 + 3H2
Mol: 0,2 0,3 0,1
PTHH: Mg + H2SO4 → MgSO4 + H2
Mol: 0,1 0,1 0,1
\(V_{ddH_2SO_4}=\dfrac{0,3+0,1}{2}=0,2\left(l\right)=200\left(ml\right)\)
\(\Rightarrow m_{ddH_2SO_4}=1,12.200=224\left(g\right)\)
c) \(C_{M_{ddAl_2\left(SO_4\right)_3}}=\dfrac{0,1}{0,2}=0,5M\)
\(C_{M_{ddMgSO_4}}=\dfrac{0,1}{0,2}=0,5M\)
\(a.2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\\ Fe+H_2SO_4\rightarrow FeSO_4+H_2\\ Đặt:\left\{{}\begin{matrix}n_{Al}=x\left(mol\right)\\n_{Fe}=y\left(mol\right)\end{matrix}\right.\\ Tacó:\left\{{}\begin{matrix}27x+56y=17,6\\1,5x+y=\dfrac{61}{112}\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}x=\dfrac{43}{190}\\y=\dfrac{2183}{10640}\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}\%m_{Al}=34,72\%\\m_{Fe}=65,28\%\end{matrix}\right.\\ b.BTNT\left(H\right):n_{H_2SO_4}=n_{H_2}=\dfrac{61}{112}\left(mol\right)\\ \Rightarrow m_{ddH_2SO_4}=\dfrac{\dfrac{61}{112}.98}{30\%}=177,92\left(g\right)\\ m_{ddsaupu}=17,6+177,92-\dfrac{61}{11,2}.2=194,43\left(g\right)\\Tacó:\left\{{}\begin{matrix}n_{AlCl_3}=\dfrac{43}{190}\\n_{FeCl_2}=\dfrac{2183}{10640}\end{matrix}\right. \\ C\%_{AlCl_3}=15,54\%;C\%_{FeCl_2}=13,4\%\)
\(a,\) Đặt \(\begin{cases} n_{Al}=x(mol)\\ n_{Fe}=y(mol \end{cases} \Rightarrow 27x+56y=17,6(1)\)
\(n_{H_2}=\dfrac{12,2}{22,4}=0,54(mol)\\ PTHH:2Al+3H_2SO_4\to Al_2(SO_4)_3+3H_2\\ Fe+H_2SO_4\to FeSO_4+H_2\\ \Rightarrow 1,5x+y=0,54(2)\\ (1)(2)\Rightarrow \begin{cases} x=0,22(mol)\\ y=0,21(mol) \end{cases} \Rightarrow \begin{cases} \%_{Al}=\dfrac{0,22.27}{17,6}.100\%=33,75\%\\ \%_{Fe}=100\%-33,75\%=66,25\% \end{cases}\\ b,\Sigma n_{H_2SO_4}=1,5x+y=0,54(mol)\\ \Rightarrow m_{dd_{H_2SO_4}}=\dfrac{0,54.98}{30\%}=176,4(g)\)
\(m_{H_2}=0,54.2=1,08(g)\\ \Rightarrow m_{dd{\text{ sau phản ứng}}}=17,6+176,4-1,08=192,92(g)\\ n_{Al_2(SO_4)_3}=\dfrac{1}{2}n_{Al}=0,11(mol);n_{FeSO_4}=n_{Fe}=0,21(mol)\\ \Sigma m_{\text{các chất sau phản ứng}}=m_{Al_2(SO_4)_3}+m_{FeSO_4}=0,11.342+0,21.152=69,54(g)\\ \Rightarrow C\%_{\text{chất sau phản ứng}}=\dfrac{69,54}{192,92}.100\%=36,05\%\)
thôi thì mình làm cho bn vậy, câu a ko làm dc đâu, làm câu b thôi, làm sao biết dc chất nào dư khi chỉ có số mol 1 chất?
nK2SO3=0.1367(mol)
mddH2SO4=Vdd.D=200.1,04=208(g)
K2SO3+H2SO4-->K2SO4+H2O+SO2
0.1367----0.1367----0.1367---------0.1367 (mol)
mddspu=100+208-0,1367.64=299.2512(g) ; mK2SO4=0,1367.174=23.7858(g)
==>C%=23.7858.100/299.512=7.94%
2)pt bn tự ghi nhé
ta có hệ pt: 56a+27b=11 và a+3b/2=8.96/22.4==>a=0.1, b=0.2
==>%Fe=0.1x56x100/11=50.9%
%Al=100%-50.9%=49.1%
b)nH2SO4= 0.7(mol)==>VddH2SO4=0.7/2=0.35(L)