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![](https://rs.olm.vn/images/avt/0.png?1311)
CuO+H2-to>Cu+H2O
0,06---0,06---0,06---0,06
n CuO=\(\dfrac{4,8}{80}=0,06mol\)
=>m Cu=0,06.64=3,84g
=>VH2=0,06.22,4=1,344l
c)Fe+2HCl->FeCl2+H2
0,06--------------------0,06 mol
=>m Fe=0,06.56=3,36g
nCuO = 4,8/80 = 0,06 (mol)
PTHH: CuO + H2 -> (t°) Cu + H2O
Mol: 0,06 ---> 0,06 ---> 0,06
mCu = 0,06.64 =3,84 (g)
VH2 = 0,06 . 22,4 = 1,344 (l)
PTHH: Fe + 2HCl -> FeCl2 + H2
Mol: 0,06 <--- 0,12 <--- 0,06 <--- 0,06
mFe = 0,06 . 56 = 3,36 (g)
mHCl = 0,12 . 36,5 = 4,38 (g)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(n_{CuO}=\dfrac{2,4}{80}=0,03\left(mol\right)\\ PTHH:CuO+H_2\rightarrow\left(t^o\right)Cu+H_2O\\ n_{H_2}=n_{Cu}=n_{CuO}=0,03\left(mol\right)\\ a,V_{H_2\left(đktc\right)}=0,03.22,4=0,672\left(l\right)\\ b,m_{Cu}=0,03.64=1,92\left(g\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(n_{Al}=\dfrac{16,2}{27}=0,6\left(mol\right)\)
2Al + 3H2SO4 ---> Al2(SO4)3 + 3H2
0,6 0,9 0,3 0,9
\(\rightarrow V_{H_2}=0,9.22,4=20,16\left(l\right)\)
\(n_{Cu}=\dfrac{57}{64}=0,890625\left(mol\right)\)
PTHH: CuO + H2 --to--> Cu + H2O
0,890625 0,890625
\(H=\dfrac{0,890625}{0,9}=99\%\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a, \(n_{H_2}=\dfrac{8,96}{22,4}=0,4\left(mol\right)\)
PTHH: 2Al + 6HCl → 2AlCl3 + 3H2
Mol: x 1,5x
PTHH: Fe + 2HCl → FeCl2 + H2
Mol: y y
b, Ta có hpt: \(\left\{{}\begin{matrix}27x+56y=11\\1,5x+y=0,4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0,2\\y=0,1\end{matrix}\right.\)
\(\Rightarrow\%m_{Al}=\dfrac{0,2.27.100\%}{11}=49,09\%\Rightarrow\%m_{Fe}=100\%-49,09\%=50,91\%\)
c, \(n_{CuO}=\dfrac{16}{80}=0,2\left(mol\right)\)
Ta có: \(\dfrac{0,2}{1}< \dfrac{0,4}{1}\) ⇒ CuO hết, H2 dư
PTHH: CuO + H2 → Cu + H2O
Mol: 0,2 0,2
\(m_{Cu}=0,2.64=12,8\left(g\right)\)
Gọi \(m_{Al}=a\left(g\right)\left(0< a< 11\right)\)
\(\rightarrow m_{Fe}=11-a\left(g\right)\)
\(\rightarrow\left\{{}\begin{matrix}n_{Al}=\dfrac{a}{27}\left(mol\right)\\n_{Fe}=\dfrac{11-a}{56}\left(mol\right)\end{matrix}\right.\)
PTHH:
\(2Al+6HCl\rightarrow2AlCl_3+3H_2\)
\(\dfrac{a}{27}\) \(\dfrac{a}{18}\)
\(Fe+2HCl\rightarrow FeCl_2+H_2\)
\(\dfrac{11-a}{56}\) \(\dfrac{11-a}{56}\)
\(\rightarrow pt:\dfrac{a}{18}+\dfrac{11-a}{56}=0,4\\ \Leftrightarrow m_{Al}=a=5,4\left(g\right)\left(TM\right)\)
\(\rightarrow\left\{{}\begin{matrix}\%m_{Al}=\dfrac{5,4}{11}=49,1\%\\\%m_{Fe}=100\%-49,1\%=50,9\%\end{matrix}\right.\)
\(n_{CuO}=\dfrac{16}{80}=0,2\left(mol\right)\)
PTHH: \(CuO+H_2\underrightarrow{t^o}Cu+H_2O\)
LTL: \(0,2< 0,4\rightarrow\) H2 dư
\(n_{Cu}=n_{CuO}=0,2\left(mol\right)\rightarrow m_{CuO}=0,2.64=12,8\left(g\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(n_{Fe}=\dfrac{2.8}{56}=0.05\left(mol\right)\)
\(Fe+2HCl\rightarrow FeCl_2+H_2\)
\(0.05................................0.05\)
\(V_{H_2}=0.05\cdot22.4=1.12\left(l\right)\)
\(n_{CuO}=\dfrac{6}{80}=0.075\left(mol\right)\)
\(CuO+H_2\underrightarrow{^{^{t^0}}}Cu+H_2O\)
\(1............1\)
\(0.075......0.05\)
Chất khử : H2 . Chất OXH : CuO
\(LTL:\dfrac{0.075}{1}>\dfrac{0.05}{1}\Rightarrow CuOdư\)
\(m_{CuO\left(dư\right)}=\left(0.075-0.05\right)\cdot64=1.6\left(g\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) nAl = 43,2/27 = 1,6 mol
2Al + 6HCl → 2AlCl3 + 3H2
1,6 \(\dfrac{1,6\times3}{2}=2,4\)
→ nH2 = 2,4 mol → VH2 = 2,4 x 22, 4 = 53,76 lít
b) nCuO = 64/80 = 0,8 mol
nH2 = 2,4 mol
→ H2 dư, phương trình tính theo số mol của CuO
CuO + H2 → Cu + H2O
0,8 0,8 0,8 0,8
Chất rắn sau phản ứng có Cu
mCu = 0,8 x 64 = 51,2 gam
![](https://rs.olm.vn/images/avt/0.png?1311)
\(n_{Zn}=\dfrac{m}{M}=\dfrac{3,25}{65}=0,05\left(mol\right)\)
\(PTHH:Zn+2HCl\rightarrow ZnCl_2+H_2\)
1 2 1 1
0,05 0,1 0,05 0,05
a) \(V_{H_2}=n.24,79=0,05.24,79=1,2395\left(l\right)\)
\(m_{ZnCl_2}=n.M=0,05.\left(65+35,5.2\right)=6,8\left(g\right)\)
b) \(PTHH:CuO+H_2\underrightarrow{t^o}Cu+H_2O\)
Ta cos tỉ lệ: \(\dfrac{0,1}{1}>\dfrac{0,05}{1}\Rightarrow\) CuO dư.
Theo ptr, ta có: \(n_{Cu}=n_{H_2}=0,05mol\\ \Rightarrow m_{Cu}=n.M=0,05.64=3,2\left(g\right).\)
a, PT: \(Zn+2HCl\rightarrow ZnCl_2+H_2\)
Ta có: \(n_{Zn}=\dfrac{3,25}{65}=0,05\left(mol\right)\)
THeo PT: \(n_{ZnCl_2}=n_{H_2}=n_{Zn}=0,05\left(mol\right)\)
\(\Rightarrow V_{H_2}=0,05.24,79=1,2395\left(l\right)\)
\(m_{ZnCl_2}=0,05.136=6,8\left(g\right)\)
b, Ta có: \(n_{CuO}=\dfrac{8}{80}=0,1\left(mol\right)\)
PT: \(CuO+H_2\underrightarrow{t^o}Cu+H_2O\)
Xét tỉ lệ: \(\dfrac{0,1}{1}>\dfrac{0,05}{1}\), ta được CuO dư.
Theo PT: \(n_{Cu}=n_{H_2}=0,05\left(mol\right)\Rightarrow m_{Cu}=0,05.64=3,2\left(g\right)\)
a) PTHH: H2 + CuO \(\underrightarrow{t^o}\) Cu + H2O(1)
b) nCuO = \(\frac{4,8}{80}=0,06\left(mol\right)\)
Theo PT(1): nCu = nCuO = 0,06 (mol)
=> mCu = 0,06.64 = 3,84 (g)
c) Theo PT(1): n\(H_2\) = nCu = 0,06 (mol) = n\(H_2\)(2)
=> V\(H_2\) = 0,06.22,4 =1,344 (l)
d) PTHH: Fe + 2HCl \(\rightarrow\) FeCl2 + H2\(\uparrow\)(2)
Theo PT(2): nFe = n\(H_2\) = 0,06 (mol)
=> mFe = 0,06.56 = 3,36 (g)
Theo PT(2): nHCl = 2n\(H_2\) = 0,12 (mol)
=> mHCl = 0,12.36,5 = 4,38(g)