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*Sửa đề: "13,44 lít H2" và "24,9 gam hh 2 kim loại"
PTHH: \(2Al+6HCl\rightarrow2AlCl_3+3H_2\uparrow\)
a_____3a_____________\(\dfrac{3}{2}\)a (mol)
\(Zn+2HCl\rightarrow ZnCl_2+H_2\uparrow\)
b_____2b_____________b (mol)
Ta lập HPT: \(\left\{{}\begin{matrix}27a+65b=24,9\\\dfrac{3}{2}a+b=\dfrac{13,44}{22,4}=0,6\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}a=0,2\\b=0,3\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}n_{Al}=0,2\left(mol\right)\\n_{Zn}=0,3\left(mol\right)\\n_{HCl}=1,2\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}m_{Al}=0,2\cdot27=5,4\left(g\right)\\m_{Zn}=19,5\left(g\right)\\m_{ddHCl}=\dfrac{1,2\cdot36,5}{7,3\%}=600\left(g\right)\end{matrix}\right.\)
nH2=0,1(mol)
PTHH: Mg + 2 HCl -> MgCl2 + H2
0,1__________0,2___________0,1(mol)
MgO + 2 HCl -> MgCl2 + H2O
0,05____0,1___0,05(mol)
mMg=0,1. 24= 2,4(g) -> mMgO=4,4-2,4= 2(g) -> nMgO=0,05((mol)
b) %mMg= (2,4/4,4).100=54,545%
=> %mMgO=45,455%
c) nHCl=0,3(mol) -> mHCl=0,3.36,5=10,95(g)
=> mddHCl=(10,95.100)/7,3=150(g)
a,
Mg+ 2HCl= MgCl2+ H2
MgO+ 2HCl= MgCl2+ H2O
b,
nH2= 2,24/22,4= 0,1 mol
=> nMg= nMgCl2= 0,5nHCl= 0,1 mol => nHCl= 0,2 mol
=> mMg= 0,1.24= 2,4g
=> mMgO= 2g
c,
nMgO= 2/40= 0,05 mol
=> nMgO= 0,5nHCl= nMgCl2= 0,05 mol
=> nHCl= 0,1 mol
Tổng lượng HCl cần dùng là 0,1+0,2=0,3 mol
=> m dd HCl= 0,3.36,5.100:7,3= 150g
\(n_{H_2}=\dfrac{2,24}{22,4}=0,1mol\)
\(Mg+2HCl\rightarrow MgCl_2+H_2\)
0,1 0,2 0,1 (mol)
\(\Rightarrow m_{Mg}=0,1\cdot24=2,4\left(g\right)\)
\(\Rightarrow m_{MgO}=4,4-2,4=2\left(g\right)\)\(\Rightarrow n_{MgO}=\dfrac{2}{40}=0,05mol\)
\(MgO+2HCl\rightarrow MgCl_2+H_2O\)
0,05 0,1
\(\Rightarrow\Sigma n_{HCl}=0,2+0,1=0,3mol\)\(\Rightarrow m_{HCl}=0,3\cdot36,5=10,95\left(g\right)\)
\(\Rightarrow m_{ddHCl}=\dfrac{10,95}{7,3}\cdot100=150\left(g\right)\)
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)
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)
\(n_{Fe} = a(mol) ; n_{Mg} = b(mol)\\ \Rightarrow 56a + 24b = 16,8 - 6,4 = 10,4(1)\\ Fe + 2HCl \to FeCl_2 + H_2\\ Mg + 2HCl \to MgCl_2 + H_2\\ n_{H_2} = a + b = \dfrac{6,72}{22,4} = 0,3(2)\)
Từ (1)(2) suy ra: a = 0,1 ; b = 0,2
Vậy :
\(\%m_{Fe} = \dfrac{0,1.56}{16,8}.100\% = 33,33\%\\ \%m_{Mg} = \dfrac{0,2.24}{16,8}.100\% = 28,57\%\\ \%m_{Cu} = 100\% - 33,33\% - 28,57\% = 38,1\%\)
Cu không tan trong HCl
\(\rightarrow m_{Cu}=4,2g\)
\(m_{Mg}=4,4-4,2=0,2g\)
\(n_{Mg}=\frac{0,2}{24}=\frac{1}{120}mol\)
PTHH: \(Mg+2HCl\rightarrow MgCl_2+H_2\)
Theo phương trình \(n_{HCl}=\frac{1}{120}.2=\frac{1}{60}mol\)
\(m_{HCl}=\frac{1}{60}.36,5=\frac{73}{120}mol\)
\(m_{ddHCl}=\frac{73}{120}:7,3\%=\frac{25}{3}g\approx8,3g\)
Theo phương trình \(n_{H_2}=n_{Mg}=\frac{1}{120}mol\)
\(V_{H_2}=\frac{1}{120}.22,4=\frac{14}{75}l\approx0,1867l\)