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\(n_{H_2}=\dfrac{2.24}{22.4}=0.1\left(mol\right)\)
\(2A+2nHCl\rightarrow2ACl_n+nH_2\)
\(\dfrac{0.2}{n}.......................0.1\)
\(M_A=\dfrac{2.4}{\dfrac{0.2}{n}}=12n\left(\dfrac{g}{mol}\right)\)
\(BL:n=2\Rightarrow M=24\)
\(A:Mg\)
\(m_{MgCl_2}=0.1\cdot95=9.5\left(g\right)\)
\(m_{ddHCl}=\dfrac{0.2\cdot36.5}{7.3\%}=100\left(g\right)\)
\(m_{dd}=2.4+100-0.1\cdot2=102.2\left(g\right)\)
\(C\%_{MgCl_2}=\dfrac{9.5}{102.2}\cdot100\%=9.3\%\)
a,Fe + 2HCl → FeCl + H2 (1)
FeO + 2HCl → FeCl + H2O (2)
nH2 = 3,36/ 22,4 = 0,15 ( mol)
Theo (1) nH2 = nFe = 0,15 ( mol)
mFe = 0,15 x 56 = 8.4 (g)
m FeO = 12 - 8,4 = 3,6 (g)
a, \(n_{H_2}=\frac{3,36}{22,4}=0,15\left(mol\right)\)
\(Fe+2HCl->FeCl_2+H_2\left(1\right)\)
\(FeO+2HCl->FeCl_2+H_2O\left(2\right)\)
theo (1) \(n_{Fe}=n_{H_2}=0,15\left(mol\right)\)
=> \(m_{Fe}=0,15.56=8,4\left(g\right)\)
=> \(m_{FeO}=12-8,4=3,6\left(g\right)\)
ta thấy : nFe =nH2 = 0,15
=> mFe =0,15 x 56 = 8,4g
%Fe=8,4/12 x 100 = 70%
=>%FeO = 100 - 70 = 30%
b) BTKLra mdd tìm mct of HCl
c) tìm mdd sau pứ -mH2 nha bạn
câu 1
Feo + 2HCl----->FeCl2 +H2
x-----------------x
ZnO + 2HCl--------->ZnCl2+ H2
y--------------------------y
\(\left\{{}\begin{matrix}72x+81y=67,95\\127x+136y=114,7\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=0,1\\y=o,75\end{matrix}\right.\)
%FeO=\(\frac{0,1.72}{67,95}\).100=10,6
%ZnO=89,4
b/ nHCl=0,1.2+0,75.2=1,7
mHcl=1,7.36,5=62,05
mddHcl=62,05.100/3,65=1700
mH2=1,7
mdd=1700+67,95-1,7=1766,25
C%Fecl2=0,1.127/1766,25.100=0,72
C%ZnCl2=0,75.136/1766,25.100=5,77
a) Mg + 2HCl -> MgCl2 + H2
Al + 3HCl -> AlCl3 + 3/2H2
b) Gọi a, b lần lượt là số mol Mg, Al.
nH2 = 5,6/22,4 = 0,25 (mol)
Mg + 2HCl -> MgCl2 + H2
a 2a a a
Al + 3HCl -> AlCl3 + 3/2H2
b 3b b 3/2b
Ta có hệ pt:
mhh = 24a + 27b = 5,1 (g)
nH2 = a + 3/2b = 0,25 (mol)
=> a = 0,1 (mol)
b = 0,1 (mol)
200 ml = 0,2 l
nHCl = 2a + 3b = 0,2 + 0,3 = 0,5 (mol)
=> CM ddHCl = 0,5/0,2 = 2,5 (M)
%mMg = 24a/5,1*100% = 2,4/5,1*100% = 47,06%
%mAl = 100%-47,06% = 52,94%
CuO + 2HCl \(\rightarrow\)CuCl2 + H2O (1)
Al2O3 + 6HCl \(\rightarrow\)2AlCl3 + 3H2O (2)
Đặt nCuO=a
nAl2O3=b
Ta có:
\(\left\{{}\begin{matrix}80a+102b=60,4\\135a+133,5.2.b=120,9\end{matrix}\right.\)
=>a=0,5;b=0,2
mCuO=80.0,5=40(g)
mAl2O3=60,4-40=20,4(g)
b;\(\sum\)nHCl=0,5.2+0,2.6=2,2(mol)
mHCl=2,2.36,5=80,3(g)
mdd HCl=80,3:\(\dfrac{14,6}{100}=550\left(g\right)\)
mCuCl2=135.0,5=67,5(g)
mAlCl3=120,9-67,5=53,4(g)
C% dd CuCl2=\(\dfrac{67,5}{60,4+550}.100\%=11,06\%\)
C% dd AlCl3=\(\dfrac{53,4}{60,4+550}.100\%=8,7\%\)
1.
Fe + 2HCl\(\rightarrow\)FeCl2 + H2
nFe=\(\dfrac{22,4}{56}=0,4\left(mol\right)\)
Theo PTHH ta có:
nH2=nFe=0,4(mol)
VH2=22,4.0,4=8,96(lít)
b;Theo PTHH ta có:
nFeCl2=nFe=0,4(mol)
nHCl=2nFe=0,8(mol)
mFeCl2=127.0,4=50,8(g)
mHCl=36,5.0,8=29,2(g)
mdd HCl=\(29,2:\dfrac{7,3}{100}=400\left(g\right)\)
C% dd FeCl2=\(\dfrac{50,8}{22,4+400}.100\%=12\%\)
2Al + 6HCl \(\rightarrow\)2AlCl3 + 3H2 (1)
Fe + 2HCl \(\rightarrow\)FeCl2 + H2 (2)
mHCl trong dd=\(175.\dfrac{7,3}{100}=12,775\left(g\right)\)
nHCl=\(\dfrac{12,775}{36,5}=0,35\left(mol\right)\)
Đặt nAl=a
nFe=b
Ta có:
\(\left\{{}\begin{matrix}27a+56b=4,1\\3a+2b=0,35\end{matrix}\right.\)
a=0,1;b=0,025
mAl=27.0,1=2,7(g)
% Al=\(\dfrac{2,7}{4,1}.100\%=65,85\%\)
%Fe=100-65,85=34,15%
b;Theo PTHH 1 và 2 ta có:
nAl=nAlCl3=0,1(mol)
nFe=nFeCl2=0,025(mol)
mAlCl3=133,5.0,1=13,35(g)
mFeCl2=0,025.127=3,175(g)
C% dd AlCl3=\(\dfrac{13,35}{4,1+175-0,175.2}.100\%=7,46\%\)
C% FeCl2=\(\dfrac{3,175}{4,1+175-0,175.2}.100\%=1,77\%\)