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Câu 1:
Gọi số mol Al là x; Zn là y
\(\rightarrow27x+65y=18,4\)
\(Al+3HCl\rightarrow AlCl_3+\frac{3}{2}H_2\)
\(Zn+2HCl\rightarrow ZnCl_2+H_2\)
\(\rightarrow n_{H2}=1,5n_{Al}+n_{Zn}=1,5x+y=\frac{1}{2}=0,5\left(mol\right)\)
Giải được: \(x=y=0,2\)
\(\Rightarrow m_{Al}=27x=5,4\left(g\right)\Rightarrow\%m_{Al}=\frac{5,4}{18,4}=29,3\%\Rightarrow\%m_{Zn}=70,7\%\)Câu 2:
\(Fe+2HCl\rightarrow FeCl_2+H_2\)
\(FeO+2HCl\rightarrow FeCl_2+H_2\)
Ta có: \(n_{H2}=n_{Fe}=\frac{2,24}{22,4}=0,1\left(mol\right)\)
Muối thu được là FeCl2
\(\rightarrow n_{FeCl2}=\frac{38,1}{56+35,5.2}=0,3\left(mol\right)\)
Ta có: \(n_{FeCl2}=n_{Fe}+n_{FeO}\rightarrow n_{FeO}=0,2\left(mol\right)\)
\(\Rightarrow m_{Fe}=0,1.56=5,6\left(g\right);m_{FeO}=0,2.\left(56+16\right)=14,4\left(g\right)\)
Câu 3 :
Cu không tác dụng với HCl, chỉ có Zn phản ứng.
\(Zn+2HCl\rightarrow ZnCl_2+H_2\)
Ta có: \(n_{H2}=\frac{4,48}{22,4}=0,2\left(mol\right)\)
Theo phản ứng: \(n_{Zn}=n_{H2}=0,2\left(mol\right)\rightarrow m_{Zn}=0,2.65=13\left(g\right)\)
\(\rightarrow\%m_{Zn}=\frac{13}{20}=65\%\rightarrow\%m_{Cu}=35\%\)
Ta có: \(n_{HCl}=2n_{H2}=0,2.2=0,4\left(mol\right)\)
\(\Rightarrow V_{HCl}=\frac{0,4}{2}=0,2\left(l\right)\)
Câu 4:
\(Fe+2HCl\rightarrow FeCl_2+H_2\)
\(Al+3HCl\rightarrow AlCl_3+\frac{3}{2}H_2\)
Gọi số mol Fe là x; Al là y
\(\rightarrow56x+27y=22\)
Ta có: \(n_{H2}=n_{Fe}=1,5n_{Al}=x+1,5y=\frac{17,92}{22,4}=0,8\left(mol\right)\)
Giải được: \(\rightarrow\left\{{}\begin{matrix}x=0,2\\y=0,4\end{matrix}\right.\)
\(\Rightarrow m_{Fe}=0,2.56=11,2\left(g\right)\)
\(\rightarrow\%m_{Fe}=\frac{11,2}{22}=50,9\%\rightarrow\%m_{Al}=49,1\%\)
Ta có: \(n_{HCl}=2n_{H2}=1,6\left(mol\right)\)
\(\rightarrow m_{HCl}=1,6.36,5=58,4\left(g\right)\)
\(\rightarrow m_{dd_{HCl}}=\frac{58,4}{7,3\%}=800\left(g\right)\)
Câu 5:
Gọi chung 2 kim loại là R hóa trị I
\(R+HCl\rightarrow RCl+\frac{1}{2}H_2\)
Ta có: \(n_{H2}=\frac{0,448}{22,4}=0,02\left(mol\right)\rightarrow n_{RCl}=2n_{H2}=0,04\left(mol\right)\)
\(\rightarrow m_{RCl}=0,04.\left(R+35,5\right)=2,58\rightarrow R=29\)
Vì 2 kim loại liên tiếp nhau \(\rightarrow\) 2 kim loại là Na x mol và K y mol
\(\rightarrow x+y=n_{RCl}=0,04\left(mol\right)\)
\(m_{hh}=m_R=23x+39y=0,04.29=1,16\left(g\right)\)
Giải được: \(\rightarrow\left\{{}\begin{matrix}x=0,025\\y=0,015\end{matrix}\right.\)
\(\rightarrow m_{Na}=0,575\left(g\right)\)
\(\rightarrow\%m_{Na}=\frac{0,575}{1,16}=49,57\%\rightarrow\%m_K=50,43\%\)
Câu 6:
Khối lượng mỗi phần là 35/2=17,5g
Gọi số mol Fe, Cu, Al là a, b, c
Ta có \(56a+64b=27c=17,5\)
Phần 1: \(n_{H2}=\frac{6,72}{22,4}=0,3\left(mol\right)\)
\(Fe+2HCl\rightarrow FeCl_2+H_2\)
\(2Al+6HCl\rightarrow2AlCl_3+3H_2\)
\(\Rightarrow a=1,5b=n_{H2}=0,3\)
Phần 2: \(n_{Cl2}=\frac{10,64}{22,4}=0,475\left(mol\right)\)
\(2Fe+3Cl_2\rightarrow2FeCl_3\)
\(Cu+Cl_2\rightarrow CuCl_2\)
\(2Al+3Cl_2\rightarrow2AlCl_3\)
\(\Rightarrow1,5a+b+1,5c=n_{Cl2}=0,465\)
\(\rightarrow\left\{{}\begin{matrix}a=0,15\\b=0,1\\c=0,1\end{matrix}\right.\)
\(\rightarrow\%m_{Fe}=\frac{0,15.56}{17,5}=48\%\)
\(\rightarrow\%m_{Cu}=\frac{0,1.64}{17,5}=36,57\%\)
\(\rightarrow\%m_{Al}=100\%-48\%-36,57\%=15,43\%\)
Câu 1
2Al+6HCl--->2Alcl3+3H2
x-----------------------1,5x
Zn+2HCl---->Zncl2+H2
y---------------------------y
n H2=1/2=0,5(mol)
Theo bài ta có hpt
\(\left\{{}\begin{matrix}27x+65y=18,4\\1,5x+y=0,5\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0,2\\y=0,2\end{matrix}\right.\)
%m Al=0,2.27/18,4.100%=29,35%
%m Zn=100%-29,35=70,65%
Câu 2.
Fe+2HCl---->FeCl2+H2
FeO+2HCl--->FeCl2+H2
n H2=2,24/22,4=0,1(mol)
m H2=0,2(g)
n Fe=n H2=0,2(mol)
m Fe=0,2.56=11,2(g)
n FeCl2(1)=2n H2=0,2(mol)
m FeCl2(1)=0,2.127=25,4(g)
m FeCl2(PT2)=38,1-25,4=12,7(g)
n FeCl2=12,7/127=0,1(mol)
n FeO=n FeCl2=0,1(mol)
m FeO=0,1.72=7,2(g)
3.
Zn+2HCl--->ZnCl2+H2
n H2=4,48/22,4=0,2(mol)
n Zn=n H2=0,2(mol)
m Zn=0,2.56=11,2(g)
%m Zn=11,2/20.100%=56%
%m Cu=100-56=34%
b) n HCl=2n H2=0,4(mol)
V H2=0,4/2=0,2(l)
4.
a) Fe+2HCl---.FeCl2+H2
x-----------------------------x(mol)
2Al+6HCl--->AlCl3+3H2
y------------------------------1,5y
n H2=17,92/22,4=0,89mol)
Theo bài ta có hpt
\(\left\{{}\begin{matrix}56x+27y=22\\x+1,5y=0,8\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0,2\\y=0,4\end{matrix}\right.\)
%m Fe=0,2.56/22.100%=50,9%
%m Al=100-50,9=49,1%
b) n HCl=2n H2=1,6(mol)
m HCl=1,6.36,5=58,4(g)
m dd HCl=58,4.100/7,3=800(g)
nH2= 0,35(mol)
a) PTHH: Mg + 2 HCl -> MgCl2 + H2
x_________2x_______x______x(mol)
PTHH: Fe + 2 HCl -> FeCl2 + H2
y________2y________y_____y(mol)
Ta có hpt: \(\left\{{}\begin{matrix}24x+56y=13,2\\x+y=0,35\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0,2\\y=0,15\end{matrix}\right.\)
b) m=m(muối khan)= mMgCl2 + mFeCl2= 95.x+127y=95.0,2+127.0,15= 38,05(g)
a)
Gọi
\(n_{Fe} = a(mol) ; n_{Mg} = b(mol)\\ \Rightarrow 56a + 24b = 13,2(1)\)
\(Mg + 2HCl \to MgCl_2 + H_2\\ Fe + 2HCl \to FeCl_2 + H_2\)
Theo PTHH : \(n_{H_2} = a + b = 0,35(mol)\)(2)
Từ (1)(2) suy ra a = 0,15 ;b = 0,2
Vậy :
\(\%m_{Fe} = \dfrac{0,15.56}{13,2}.100\% = 63,64\%\\ \Rightarrow m_{Mg} = 100\% - 63,64\% = 36,36\%\)
b)
Ta có :\(n_{HCl} = 2n_{H_2} = 0,7(mol)\)
Bảo toàn khối lượng :
\(m_{muối} = m_{kim\ loại} + m_{HCl} - m_{H_2} = 13,2 + 0,7.36,5 - 0,35.2=38,05(gam)\)
Sửa đề: đktc → đkc
a, \(Mg+2HCl\rightarrow MgCl_2+H_2\)
\(Fe+2HCl\rightarrow FeCl_2+H_2\)
Ta có: 24nMg + 56nFe = 13,2 (1)
\(n_{H_2}=\dfrac{8,6765}{24,79}=0,35\left(mol\right)\)
Theo PT: \(n_{H_2}=n_{Mg}+n_{Fe}=0,35\left(2\right)\)
Từ (1) và (2) \(\Rightarrow\left\{{}\begin{matrix}n_{Mg}=0,2\left(mol\right)\\n_{Fe}=0,15\left(mol\right)\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\%m_{Mg}=\dfrac{0,2.24}{13,2}.100\%\approx36,36\%\\\%m_{Fe}\approx63,64\%\end{matrix}\right.\)
b, Theo PT: \(\left\{{}\begin{matrix}n_{MgCl_2}=n_{Mg}=0,2\left(mol\right)\\n_{FeCl_2}=n_{Fe}=0,15\left(mol\right)\end{matrix}\right.\)
⇒ m muối khan = 0,2.95 + 0,15.127 = 38,05 (g)
\(Đặt:n_{MnO_2}=a\left(mol\right),n_{KMnO_4}=b\left(mol\right)\)
\(m_{hh}=87a+158b=37.96\left(g\right)\left(1\right)\)
\(n_{Cl_2}=\dfrac{10.08}{22.4}=0.45\left(mol\right)\)
\(2KMnO_4+16HCl\rightarrow2KCl+2MnCl_2+5Cl_2+8H_2O\)
\(MnO_2+4HCl\rightarrow MnCl_2+Cl_2+2H_2O\)
\(n_{Cl_2}=a+2.5b=0.45\left(mol\right)\left(2\right)\)
\(\left(1\right),\left(2\right):a=0.4,b=0.02\)
\(\%MnO_2=\dfrac{0.4\cdot87}{37.96}\cdot100\%=91.68\%\\\%KMnO_4=100-91.68=8.32\% \)
\(m_M=m_{KCl}+m_{MnCl_2}=0.02\cdot74.5+\left(0.4+0.02\right)\cdot126=54.41g\)
Đặt :
nFe = x mol
nMgO = y mol
mX = 56x + 40y = 13.6 (g) (1)
Fe + 2HCl => FeCl2 + H2
x____________x
MgO + 2HCl => MgCl2 + H2O
y______________y
mM = mFeCl2 + mMgCl2 = 127x + 95y = 31.7 (2)
(1) , (2) :
x = 0.1
y = 0.2
%Fe = 5.6/13.6 * 100% = 41.17%
%MgO = 58.82%
nKOH = 0.1 * 0.2 = 0.02 (mol)
KOH + HCl => KCl + H2O
0.02____0.02
nHCl (pư) = 2nFe + 2nMgO = 0.1*2 + 0.2*2 = 0.6 (mol)
nHCl = 0.02 + 0.6 = 0.62 (mol)
VddHCl = 0.62/0.5 = 1.24 (M)