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![](https://rs.olm.vn/images/avt/0.png?1311)
a, PT: \(Zn+2HCl\rightarrow ZnCl_2+H_2\)
\(Fe+2HCl\rightarrow FeCl_2+H_2\)
b, Ta có: \(n_{HCl}=0,5.0,2=0,1\left(mol\right)\)
\(n_{H_2}=\dfrac{0,896}{22,4}=0,04\left(mol\right)\)
Theo PT: \(n_{HCl\left(pư\right)}=2n_{H_2}=0,08\left(mol\right)< 0,1\left(mol\right)\)
→ HCl dư.
Gọi: \(\left\{{}\begin{matrix}n_{Zn}=x\left(mol\right)\\n_{Fe}=y\left(mol\right)\end{matrix}\right.\)
Theo PT: \(n_{H_2}=n_{Zn}+n_{Fe}=x+y=0,04\left(1\right)\)
\(\left\{{}\begin{matrix}n_{ZnCl_2}=n_{Zn}=x\left(mol\right)\\n_{FeCl_2}=n_{Fe}=y\left(mol\right)\end{matrix}\right.\)⇒ 136x + 127y = 5,26 (2)
Từ (1) và (2) \(\Rightarrow\left\{{}\begin{matrix}x=0,02\left(mol\right)\\y=0,02\left(mol\right)\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\%m_{Zn}=\dfrac{0,02.65}{0,02.65+0,02.56}.100\%\approx53,72\%\\\%m_{Fe}\approx46,28\%\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a, \(Zn+2HCl\rightarrow ZnCl_2+H_2\)
b, \(n_{Zn}=\dfrac{13}{65}=0,2\left(mol\right)\)
Theo PT: \(n_{H_2}=n_{Zn}=0,2\left(mol\right)\Rightarrow V_{H_2}=0,2.22,4=4,48\left(l\right)\)
c, \(n_{ZnCl_2}=n_{Zn}=0,2\left(mol\right)\Rightarrow m_{ZnCl_2}=0,2.136=27,2\left(g\right)\)
d, \(n_{HCl}=2n_{Zn}=0,4\left(mol\right)\Rightarrow m_{HCl}=0,4.36,5=14,6\left(g\right)\)
\(\Rightarrow m_{ddHCl}=\dfrac{14,6}{7,3\%}=200\left(g\right)\)
⇒ m dd sau pư = 13 + 200 - 0,2.2 = 212,6 (g)
\(\Rightarrow C\%_{ZnCl_2}=\dfrac{27,2}{212,6}.100\%\approx12,79\%\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a, \(MgO+H_2SO_4\rightarrow MgSO_4+H_2O\)
b, Ta có: \(m_{H_2SO_4}=200.9,8\%=19,6\left(g\right)\)
\(\Rightarrow n_{H_2SO_4}=\dfrac{19,6}{98}=0,2\left(mol\right)\)
Theo PT: \(n_{MgO}=n_{MgSO_4}=n_{H_2SO_4}=0,2\left(mol\right)\)
\(\Rightarrow m_{MgO}=0,2.40=8\left(g\right)\)
c, Ta có: m dd sau pư = 8 + 200 = 208 (g)
\(\Rightarrow C\%_{MgSO_4}=\dfrac{0,2.120}{208}.100\%\approx11,54\%\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(n_{H_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\\ pthh:Mg+2HCl\rightarrow MgCl_2+H_2\)
0,2 0,2
\(MgO+2HCl\rightarrow MgCl_2+H_2O\)
\(m_{Mg}=0,2.24=4,8g\\ m_{MgO}=18,4-4,8=13,6g\)
![](https://rs.olm.vn/images/avt/0.png?1311)
ta có phương trình:
2Al+6HCl=>2AlCl3+3H2
a 1.5a
Mg+2HCl=>MgCl2+H2
b b
ta có vH2=13.44(lít)=>nH2=13.4422.4=0.6(mol)
gọi a là số mol của Al,b là số mol của Mg
=>1.5a+b=0.6(mol)(1)
27a+24b=12.6(g)(2)
từ (1)(2)=>a=0.2(mol),b=0.3(mol)
=>%Al=0.2∗2712.6*100=42.86%
=>%Mg=100-42.86=57.14%
2KMnO4--->K2MnO4+MnO2+O2 n KMnO4=15,8/158=0,1(mol) n O2=1/2n KMnO4=0,05(mol) V O2=0,05.22,4=1,12(l)
Câu 2: a) SO3+H2O--->H2SO4 b) m H2SO4=20.10/100=2(g) n H2SO4=2/98=0,02(mol) n SO3=n H2SO4=0,02(mol) m =m SO3=0,02.80=1,6(g)
Câu 3 : a) Fe+2HCl-->FeCl2+H2 x--------------------------x(mol) Mg+2HCl------->MgCl2+H2 y------------------------------y(mol) n H2=4,48/22,4=0,2(mol) Theo bài ra ta có hpt \(\left\{{}\begin{matrix}56x+24y=8\\x+y=0,2\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0,1\\y=0,1\end{matrix}\right.\) n Fe : n Al= 1 : 1
Câu 4: a) Hiện tượng : có chất rắn màu nâu đỏ sau pư PT: FeCl3+3KOH--->3KCl+Fe(OH)3 b) m KOH=\(\frac{200.8,4}{100}=16,8\left(g\right)\) n KOH=16,8/56=0,3(mol) n Fe(OH)3=1/3n KOH=0,1(mol) m Fe(OH)3=0,1.107=10,7(g) c) n FeCl3=1/3n KOH=0,1(mol) m FeCl3=0,1.162,5=16,25(g) m dd FeCl3=16,25.100/6,5=250(g) m dd sau pư=m FeCl3+m dd KOH- m Fe(OH)3 =250+200-10,7=439,3(g) n KCl=n KOH=0,3(mol) m KCl=74,5.0,3=22,35(g) C% KCl=22,35/439,3.100%=5,09% d) 2Fe(OH)3--->Fe2O3+3H2O n Fe2O3=1/2n Fe(OH)3=0,05(mol) a=m Fe2O3=0,05.160=8(g)
Câu 5: a) Al2O3+6HCl---->2Alcl3+3H2O x----------6x(mol) MgO+2HCl----->MgCl2+H2O y-----------2y(mol) m HCl=90.7,3/100=6,57(g) n HCl=6,57/36,5=0,18(mol) Theo bài ta có hpt \(\left\{{}\begin{matrix}102x+40y=3,24\\6x+2y=0,18\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0,02\\y=0,03\end{matrix}\right.\) %m Al2O3=\(\frac{0,02.102}{3,24}.100\%=62,96\%\) %m MgO=100-62,96=37,04% b)m dd sau pư=m KL+m dd HCl=3,24+90=93,24(g) m AlCl3=0,04.133,5=5,34(g) C% Alcl3=5,34/92,24.100%=5,79% m MgCl2=0,03.95=2,85(g) C% MgCl2=2,85/92,24.100%=2,8%
Câu 6: a) n H2=1,456/22,4=0,056(mol) 2Al+3H2SO4---.Al2(SO4)3+3H2 x-------------------------------1,5x Fe+H2SO4--->FeSO4+H2 y--------------------------------y(mol) Theo bài ra ta có hpt \(\left\{{}\begin{matrix}27x+56y=1,93\\1,5x+y=0,065\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0,03\\y=0,02\end{matrix}\right.\) %m Al=0,03.27/1,93.100%=41,97% %m Fe=100-41,97=53,08% c) 2Al+3Cu(NO3)2---->3Cu+2Al(NO3) 0,03------------------------0,045(mol) Fe+Cu(NO3)2---->Fe(NO3)2+Cu 0,02-------------------------------0,02(mol) m Cu=(0,045+0,02).64=4,16(g)
Câu 7: a) Zn+2HCl---.Zncl2+H2 b) n H2=3,36/22,4=0,15(mol) n Zn=n H2=0,15(mol) m Zn=0,15.65=9,75(g) %m Zn=9,75/10,05.100%=97% %m Cu=3%
Câu 9:
Gọi oxit KL Cần tìm là MO
MO+H2SO4--->MSO4+H2O
n H2SO4=19,6/98=0,2(mol)
n MO=n H2SO4=0,2(mol)
M MO=16/0,2=80
M+18=80-->M=64(Cu)
Vậy M là Cu
M chơi người à Dương