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a) \(n_{PbS}=\dfrac{23,9}{239}=0,1\left(mol\right)\)
=> \(n_{H_2S}=0,1\left(mol\right)\)
\(\%V_{H_2S}=\dfrac{0,1.22,4}{2,464}.100\%=90,9\%\)
\(\%V_{H_2}=100\%-90,9\%=9,1\%\)
b) \(n_{H_2}=\dfrac{2,464.9,1\%}{22,4}=0,01\left(mol\right)\)
PTHH: Fe + 2HCl --> FeCl2 + H2
0,01<-------------------0,01
FeS + 2HCl --> FeCl2 + H2S
0,1<---------------------0,1
=> \(\left\{{}\begin{matrix}\%m_{Fe}=\dfrac{0,01.56}{0,01.56+0,1.88}.100\%=5,983\%\\\%m_{FeS}=\dfrac{0,1.88}{0,01.56+0,1.88}.100\%=94,017\%\end{matrix}\right.\)
Đặt \(n_{FeS}=a\left(mol\right);n_{Na_2S}=b\left(mol\right)\)
\(\left(1\right)FeS+2HCl\rightarrow FeCl_2+H_2S\)
(mol)_a______2a_____a________a_
\(\left(2\right)Na_2S+2HCl\rightarrow2NaCl+H_2S\)
(mol)__b_____2b_______2b_____b__
Theo đề ta có hpt:
\(\left\{{}\begin{matrix}88a+78b=25,4\\36,5\left(2a+2b\right)=\frac{60.36,5}{100}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=0,2\\b=0,1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\%m_{FeS}=\frac{0,2.88}{25,4}.100\%=69,3\left(\%\right)\\\%m_{Na_2S}=100-69,3=30,7\left(\%\right)\end{matrix}\right.\)
\(PTHH:FeCl_2+Pb\left(NO_3\right)_2\rightarrow PbCl_2+Fe\left(NO_3\right)_2\)
(mol)_____0,2___________________0,2______________
\(PTHH:2NaCl+Pb\left(NO_3\right)_2\rightarrow PbCl_2+2NaNO_3\)
(mol)______0,2__________________0,1____________
\(m_{\downarrow}=\left(0,2+0,1\right).278=83,4\left(g\right)\)
\(\text{Đặt }n_{FeS}=a\left(mol\right);n_{Na_2S}=b\left(mol\right)\\ \Rightarrow m_{h^2}=88a+78b=25,4\left(1\right)\\ BTNT.S\Rightarrow n_{S^{2-}}=n_{Na_2S}+n_{FeS}=a+b\left(mol\right)\\ n_{HCl}=0,6\left(mol\right)\)
Vai trò H+ : 2H+ + S2- ---> H2S
__________0,6___0,3
\(\Rightarrow n_{S^{2-}}=a+b=0,3\left(2\right)\)
Từ (1) và (2) \(\Rightarrow\left\{{}\begin{matrix}a=0,2\\b=0,1\end{matrix}\right.\)
\(\Rightarrow m_{FeS}=17,6\left(g\right);m_{Na_2S}=7,8\left(g\right)\\ \Rightarrow\%FeS=69,29\%;\%Na_2S=30,71\%\)
b) BTNT.Cl \(\Rightarrow n_{PbCl_2}=\frac{1}{2}n_{HCl}=0,3\left(mol\right)\)
\(\Rightarrow m_{PbCl_2}=83,4\left(g\right)\)
\(n_{HCl} = \dfrac{448.1,12.3,65\%}{36,5} = 0,50176(mol)\\ Zn + 2HCl \to ZnCl_2 + H_2\\ ZnO + 2HCl \to ZnCl_2 + H_2O\\ n_{Zn} = n_{H_2} = \dfrac{2,24}{22,4} = 0,1(mol)\\ n_{ZnO} = \dfrac{n_{HCl} - 2n_{Zn}}{2} = \dfrac{0,50176-0,1.2}{2} = 0,15088(mol)\\ \%m_{Zn} = \dfrac{0,1.65}{0,1.65 + 0,15088.81}.100\% = 34,72\%\\ \%m_{ZnO} = 65,28\%\)
\(H_2S + Pb(NO_3)_2 \to PbS + 2HNO_3\\ n_{H_2S} = n_{PbS} = \dfrac{23,9}{239} = 0,1(mol)\\ \Rightarrow n_{H_2} = \dfrac{4,48}{22,4} - 0,1 = 0,1(mol)\\ Fe + 2HCl \to FeCl_2 + H_2\\ FeS + 2HCl \to FeCl_2 + H_2S\\ Fe + S \xrightarrow{t^o} FeS\\ n_{Fe} = n_{Fe} + n_{FeS} = n_{H_2} + n_{H_2S} = 0,2(mol)\\ n_S = n_{FeS} = n_{H_2S} = 0,1(mol)\\ \Rightarrow m = 0,2.56 + 0,1.32 = 14,4(gam) \)
Gọi số mol Na, Zn là a, b
=> 23a + 65b = 14,3
n H2=\(\dfrac{2,24}{22,4}\)=0,1 mol
- Nếu Zn tan hết
2Na + 2H2O --> 2NaOH + H2
a-------------------->a---->0,5a
2NaOH + Zn --> Na2ZnO2 + H2
2b<----b-------------------->b
->\(\left\{{}\begin{matrix}2b\text{≤}a\\0,5a+b=14,3\end{matrix}\right.\) loại
=> Zn không tan hết => NaOH hết
PT:2Na + 2H2O --> 2NaOH + H2
a------------------->a---->0,5a
2NaOH + Zn --> Na2ZnO2 + H2
\ a--------------------------->0,5a
=> 0,5a + 0,5a = 0,1
=> a = 0,1
=> mNa = 0,1.23 = 2,3 (g)
=> mZn = 14,3 - 2,3 = 12(g)
PTP. ứng:
Fe+S----> FeS
Fe+ 2HCl--> FeCl2+H2
x x (mol)
FeS+ 2HCl--> FeCl2+ H2S
y y (mol)
Gọi n H2 và H2S trong C lần lượt là x và y mol, ta có hpt:
x+y=0.1
(2x+32y)/2(x+y)=13
--> x=0.02, y=0.08
--> tổng m Fe+n FeS= n Fe tỏng A=x+y=0.1 mol
m Fe=0.1.56=5.6 (g)
\(n_{khi}=n_{H2}=\frac{11,2}{22,4}=0,5\left(mol\right)\)
Gọi số mol Mg, Fe, Cu là a, b, c
Ta có \(24a+56b+64c=28\)
\(Mg+2HCl\rightarrow MgCl_2+H_2\)
\(Fe+2HCl\rightarrow FeCl_2+H_2\)
\(n_{H2}=n_{Mg}+n_{Fe}=a+b=0,5\)
Cu không phản ứng
\(\rightarrow64c=9,6\)
\(\rightarrow\left\{{}\begin{matrix}a=0,3\\b=0,2\\c=0,15\end{matrix}\right.\)
Cho tác dụng với NaOH
\(HCl+NaOH\rightarrow NaCl+H_2O\)
\(MgCl_2+2NaOH\rightarrow Mg\left(OH\right)_2+2NaCl\)
0,3 _______________0,3_______________
\(FeCl_2+2NaOH\rightarrow Fe\left(OH\right)_2+2NaCl\)
0,2 ________________0,2_______________
\(Mg\left(OH\right)_2\rightarrow MgO+H_2O\)
0,3___________0,3_______
\(Fe\left(OH\right)_2\rightarrow FeO+H_2O\)
0,2__________0,2_______
\(\rightarrow m=0,3.40+0,2.72=26,4\left(g\right)\)
\(n_{HCl}=\frac{60.36,5\%}{36,5}=0,6\left(mol\right)\)
\(FeS+2HCl\rightarrow FeCl_2+H_2S\)
\(Na_2S+2HCl\rightarrow2NaCl+H_2S\)
Gọi a là mol FeS; b là mol Na2S
\(\left\{{}\begin{matrix}88a+78b=25,4\\2a+2b=0,6\rightarrow\end{matrix}\right.\left\{{}\begin{matrix}a=0,2\\b=0,1\end{matrix}\right.\)
\(\%_{FeS}=\frac{0,2.88.100}{25,4}=69,29\%\)
\(\%_{Na2S}=100\%-69,29\%=30,71\%\)
\(n_{FeCl2}=0,2\left(mol\right);n_{NaCl}=0,2\left(mol\right)\)
\(\rightarrow n_{Cl}=2n_{FeCl2}+n_{NaCl}=0,6\left(mol\right)\)
\(PTHH:Pb+2Cl\rightarrow PbCl_2\)
\(\rightarrow n_{PbCl2}=0,3\left(mol\right)\)
\(\rightarrow m+PbCl2=83,4\left(g\right)\)
Chỗ dòng cuối là dấu mPbCl2 = 83,2(g) nha