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![](https://rs.olm.vn/images/avt/0.png?1311)
2Al+6HCl->2AlCl3+3H2
1,2------------------0,6 mol
H2+CuO->Cu+H2O
0,4----0,4
m HCl=43,8=>n HCl=\(\dfrac{43,8}{36,5}\)=1,2 mol
=>VH2=0,6.22,4=13,44l
b)n CuO=\(\dfrac{32}{80}\)=0,4 mol
=>H2 dư
=>m=m Cu=0,4.64=25,6g
=>%mCu=100%
![](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: \(n_{Cu}=n_{H_2}=0,05\left(mol\right)\Rightarrow m_{Cu}=0,05.64=3,2\left(g\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) 2Al+6HCl→2AlCl3+3H22Al+6HCl→2AlCl3+3H2
b) nAl=5,427=0,2(mol)nAl=5,427=0,2(mol)
Theo phương trình : nH2=32nAl=0,3(mol)nH2=32nAl=0,3(mol)
→VH2(đktc)=0,3.22,4=6,72(l)→VH2(đktc)=0,3.22,4=6,72(l)
c) Chất rắn : 0,2(mol)0,2(mol)
CuO dư : 0,2(mol)Cu0,2(mol)Cu
%CuO=0,2.80(0,2.80+0,2.64).100=55,56%%CuO=0,2.80(0,2.80+0,2.64).100=55,56%
%Cu=44,44%%Cu=44,44%
a)\(n_{Al}=\dfrac{5,4}{27}=0,2mol\)
\(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\)
0,2 0,3 0,1 0,3
b)\(V_{H_2}=0,3\cdot22,4=6,72l\)
c)\(n_{CuO}=\dfrac{32}{80}=0,4mol\)
\(CuO+H_2\rightarrow Cu+H_2O\)
0,4 0,3 0,3
\(m_{Cu}=0,3\cdot64=19,2g\)
![](https://rs.olm.vn/images/avt/0.png?1311)
nAl = 5.4/27 = 0.2 (mol)
2Al + 6HCl => 2AlCl3 + 3H2
0.2.......0.6......................0.3
CM HCl = 0.6 / 0.4 = 1.5 (M)
nCuO = 32/80 = 0.4 (mol)
CuO + H2 -to-> Cu + H2O
0.2.......0.2..........0.2
Chất rắn : 0.2 (mol) CuO dư , 0.2 (mol) Cu
%CuO =\(\dfrac{0,2.80}{0,2.80+0,2.64}\) 100% = 55.56%
%Cu = 44.44%
a) \(2Al+6HCl\rightarrow2AlCl_3+3H_2\)
b) \(n_{Al}=\dfrac{5,4}{27}=0,2\left(mol\right)\)
Theo phương trình : \(n_{H_2}=\dfrac{3}{2}n_{Al}=0,3\left(mol\right)\)
\(\rightarrow V_{H_2}\left(đktc\right)=0,3.22,4=6,72\left(l\right)\)
c) Chất rắn : \(0,2\left(mol\right)\)
CuO dư : \(0,2\left(mol\right)Cu\)
\(\%CuO=\dfrac{0,2.80}{\left(0,2.80+0,2.64\right)}.100=55,56\%\)
\(\%Cu=44,44\%\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(n_{Al}=\dfrac{8,1}{27}=0,3(mol);n_{HCl}=\dfrac{21,9}{36,5}=0,6(mol)\\ a,PTHH:2Al+6HCl\to 2AlCl_3+3H_2\\ LTL:\dfrac{0,3}{2}>\dfrac{0,6}{6}\Rightarrow Al\text { dư}\\ n_{Al(dư)}=0,3-\dfrac{0,6}{3}=0,1(mol)\\ \Rightarrow m_{Al(dư)}=0,1.27=2,7(g)\\ c,n_{AlCl_3}=\dfrac{1}{3}n_{HCl}=0,2(mol)\\n_{H_2}=\dfrac{1}{2}n_{HCl}=0,3(mol)\\ \Rightarrow m_{AlCl_3}=0,2.133,5=26,7(g)\\ d,V_{H_2}=0,3.22,4=6,72(l)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a) Zn + H_2SO_4 \to ZnSO_4 + H_2\\ n_{Zn} = \dfrac{13}{65} = 0,2 < n_{H_2SO_4} = \dfrac{200.24,5\%}{98} = 0,5 \to H_2SO_4\ dư\\ n_{H_2SO_4\ pư} =n_{Zn} = 0,2(mol)\\ \Rightarrow m_{H_2SO_4\ dư} = (0,5 - 0,2).98 = 29,4(gam)\\ c) n_{FeSO_4} = n_{H_2} = n_{Zn} = 0,2(mol)\\ m_{FeSO_4} = 0,2.152 = 30,4(gam)\\ V_{H_2} = 0,2.22,4 = 4,48(lít)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) nAl=0,2(mol)
PTHH: 2 Al + 6 HCl -> 2 AlCl3 + 3 H2
H2 + CuO -to-> Cu + H2O
nAlCl3= nAl= 0,2(mol)
=> mAlCl3= 133,5. 0,2= 26,7(g)
b) nCu= nH2= 3/2 . 0,2=0,3(mol)
=> mCu= 0,3.64=19,2(g)
(Qua phản ứng nghe kì á, chắc tạo thành chứ ha)
<3
![](https://rs.olm.vn/images/avt/0.png?1311)
CuO+H2to→Cu+H2O
Theo PT: nCuO=nCu(1)
Ta có mrắngiảm=mCuO−mCu=3,2(g)
→80nCuO−64nCu=3,2(2)
Từ (1)(2)→nCuO=nCu=\(\dfrac{3,2}{80-64}\)=0,2(mol)(1)(2)
→nCuO=nCu=3,280−64=0,2(mol)
Theo PT: nH2=nCu=0,2(mol)
Đặt hóa trị R là n(n>0)
2R+2nHCl→2RCln+nH2
Theo PT: nR.n=2nH2
→\(\dfrac{13n}{MR}\)=0,4
→MR=32,5n
Với n=2→MR=65(g/mol)
→R là kẽm (Zn)
\(nAl=\dfrac{8,1}{27}=0,3\left(mol\right)\)
\(nHCl=\dfrac{21,9}{36,5}=0,6\left(mol\right)\)
PTHH:
\(2Al+6HCl\rightarrow2AlCl_3+3H_2\)
2 6 2 3 (mol)
0,2 0,6 0,2 0,3 (mol)
LTL : 0,3 / 2 > 0,6/6
=> Al dư sau pứ , HCl đủ vs pứ
\(mAl_{\left(dư\right)}=\left(0,3-0,2\right).27=2,7\left(g\right)\)
\(mAlCl_3=0,2.98=19,6\left(g\right)\)
\(H_2+CuO\rightarrow Cu+H_2O\)
1 1 1 1 (mol)
0,3 0,3 0,3 0,3 (mol)
=> \(mCu=0,3.64=19,2\left(g\right)\)
\(n_{Al}=\dfrac{8,1}{27}=0,3\left(mol\right)\\ n_{HCl}=\dfrac{21,9}{36,5}=0,6\left(mol\right)\\ pthh:2Al+6HCl\rightarrow2AlCl_3+3H_2\uparrow\)
\(LTL:\dfrac{0,3}{2}>\dfrac{0,6}{6}\)
=> Al dư HCl hết
theo pthh : \(n_{Al\left(p\text{ư}\right)}=\dfrac{1}{3}n_{HCl}=0,2\left(mol\right)\\ m_{Al\left(d\right)}=\left(0,3-0,2\right).27=2,7\left(g\right)\)
theo pthh : \(n_{AlCl_3}=\dfrac{1}{3}n_{HCl}=0,1\left(mol\right)\\ m_{AlCl_3}=0,1.133,5=13,35\left(g\right)\)
theo pthh : \(n_{H_2}=\dfrac{1}{2}n_{HCl}=0,3\left(mol\right)\)
pthh: \(CuO+H_2\underrightarrow{t^o}H_2O+Cu\)
0,3 0,3
\(m_{Cu}=0,3.64=19,2\)