cho 26g zn + hcl 10% a, pt? b, mznhcl2 và vh2 c, tính C% của các chất trong dd giúp e với mn ơi gấp lắm ạ huhu
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![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
Zn+2HCl->ZnCl2+H2
0,4---0,8-----0,4----0,4
n HCl=0,9 mol
n Zn=0,4 mol
=>HCl dư
=>VH2=0,4.22,4=8,96l
=> m muối=0,4.136=54,4g
![](https://rs.olm.vn/images/avt/0.png?1311)
a) $n_{Zn} = \dfrac{26}{65} = 0,4(mol)$
$Zn + 2HCl \to ZnCl_2 + H_2$
Theo PTHH : $n_{HCl} = 2n_{Zn} = 0,8(mol) \Rightarrow V_{dd\ HCl} = \dfrac{0,8}{1} = 0,8(lít)$
b)
$ZnCl_2 + 2AgNO_3 \to 2AgCl + Zn(NO_3)_2$
$n_{AgNO_3} = 2n_{ZnCl_2} = 0,8(mol)$
$V_{dd\ AgNO_3} = \dfrac{0,8}{0,5} = 1,6(lít)$
c)
$n_{AgCl} = n_{AgNO_3} = 0,8(mol) \Rightarrow m_{AgCl} = 0,8.143,5 = 114,8(gam)$
![](https://rs.olm.vn/images/avt/0.png?1311)
a)
$Zn + S \xrightarrow{t^o} ZnS$
$n_{Zn} =\dfrac{9,75}{65} = 0,15 > n_S = \dfrac{3,84}{32} = 0,12$ nên Zn dư
$n_{ZnS} = n_S = 0,12(mol)$
$m_{ZnS} = 0,12.97 = 11,64(gam)$
$n_{Zn\ dư} = 0,15 - 0,12 = 0,03(mol)$
$m_{Zn\ dư} = 0,03.65 = 1,95(gam)$
b)
$Zn + 2HCl \to ZnCl_2 + H_2$
$ZnS + 2HCl \to ZnCl_2 + H_2S$
$n_{khí} = n_{H_2} + n_{H_2S} = n_{Zn\ dư} + n_{ZnS} = 0,15(mol)$
$V = 0,15.22,4 = 3,36(lít)$
\(n_{Zn}=\dfrac{9.75}{65}=0.15\left(mol\right)\)
\(n_S=\dfrac{3.84}{32}=0.12\left(mol\right)\)
\(Zn+S\underrightarrow{^{^{t^0}}}ZnS\)
Lập tỉ lệ :
\(\dfrac{0.15}{1}>\dfrac{0.12}{1}\Rightarrow Zndư\)
\(a.\)
\(m_X=m_{ZnS}+m_{Zn\left(dư\right)}=0.12\cdot97+\left(0.15-0.12\right)\cdot65=13.59\left(g\right)\)
\(b.\)
\(Zn+2HCl\rightarrow ZnCl_2+H_2\)
\(0.03..................................0.03\)
\(ZnS+2HCl\rightarrow ZnCl_2+H_2S\)
\(0.12.................................0.12\)
\(V_{khí}=0.03\cdot22.4+0.12\cdot22.4=3.36\left(l\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 1:
n HCl = 2.0,15 = 0,3 (mol)
Zn + 2HCl --> ZnCl2 + H2
0,15 <-- 0,3 --> 0,15 (mol)
m Zn = 0,15.65 = 9,75 (g)
V H2 = 0,15.22,4 = 3,36 (l)
Bài 2:
n Al = 15/27 = 5/9 (mol)
4Al + 3O2 --> 2Al2O3
5/9 --> 5/12 --> 5/18 (mol)
m Al2O3 = 5/18 . 102 = 85/3 (g)
V O2 cần = 5/12 . 22,4 = 28/3 (l)
=> V kk cần = 28/3 .100/20 = 140/3 (l)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(n_{Zn}=\dfrac{13}{65}=0,2\left(mol\right)\\Zn+2HCl\rightarrow ZnCl_2+H_2\\ n_{HCl}=0,2.2=0,4\left(mol\right)\\ m_{ddHCl}=\dfrac{0,4.36,5.100}{10}=146\left(g\right)\\ n_{H_2}=n_{ZnCl_2}=n_{Zn}=0,2\left(mol\right)\\ V_{H_2\left(đktc\right)}=0,2.22,4=4,48\left(l\right)\\ C\%_{ddZnCl_2}=\dfrac{136.0,2}{13+146-0,2.2}.100\approx17,15\%\)
\(n_{Zn}=\dfrac{13}{65}=0,2\left(mol\right)\)
PTHH: Zn + 2HCl → ZnCl2 + H2
Mol: 0,2 0,4 0,2 0,2
\(V_{H_2}=0,2.22,4=4,48\left(l\right)\)
\(m_{ddHCl}=\dfrac{0,4.36,5.100}{10}=146\left(g\right)\)
mdd sau pứ = 13+146-0,2.2 = 158,6 (g)
\(C\%_{ddZnCl_2}=\dfrac{0,2.136.100\%}{158,6}=17,15\%\)
![](https://rs.olm.vn/images/avt/0.png?1311)
PTHH:
Zn + 2HCl ---> ZnCl2 + H2
\(n_{H2}=\frac{6,72}{22,4}=0,3\left(mol\right)\)( mình làm theo đktc nhé)
Theo PTHH:
\(n_{H2}=n_{Zn}=n_{ZnCl2}=0,3\left(mol\right)\)
a, \(m_{ZnCl2}=0,3\cdot\left(65+35,5\cdot2\right)=40,8\left(g\right)\)
b,
\(a=m_{Zn}=0,3\cdot65=19,5\left(g\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
PT: \(Zn+2HCl\rightarrow ZnCl_2+H_2\)
\(ZnO+2HCl\rightarrow ZnCl_2+H_2O\)
Ta có: \(n_{H_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\)
Theo PT: \(n_{Zn}=n_{H_2}=0,2\left(mol\right)\)
\(\Rightarrow m_{Zn}=0,2.65=13\left(g\right)\)
\(m_{ZnO}=21,1-13=8,1\left(g\right)\)
Có: \(n_{ZnO}=\dfrac{8,1}{81}=0,1\left(mol\right)\)
Theo PT: \(n_{HCl}=2n_{Zn}+2n_{ZnO}=0,6\left(mol\right)\)
\(\Rightarrow m_{HCl}=0,6.36,5=21,9\left(g\right)\Rightarrow C\%_{ddHCl}=\dfrac{21,9}{200}.100\%=10,95\%\)
Theo PT: \(n_{ZnCl_2}=n_{Zn}+n_{ZnO}=0,3\left(mol\right)\)
\(\Rightarrow m_{ZnCl_2}=0,3.136=40,8\left(g\right)\)
Bạn tham khảo nhé!
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![](https://rs.olm.vn/images/avt/0.png?1311)
- Cả 2 chất trong hhA đều tác dụng được với dd HCl dư. Nhưng chỉ có Zn tác dụng với dd HCl dư mới sinh ra khí H2
\(n_{H_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\\ PTHH:\left(1\right)Zn+2HCl\rightarrow ZnCl_2+H_2\\ \left(2\right)ZnO+2HCl\rightarrow ZnCl_2+H_2O\\ TheoPTHH\left(1\right):n_{Zn}=n_{ZnCl_2\left(1\right)}=n_{H_2}=0,2\left(mol\right)\\ m_{ZnO}=m_{hhA}-m_{Zn}=21,1-65.0,2=8,1\left(g\right)\\ n_{ZnO}=\dfrac{8,1}{81}=0,1\left(mol\right)\\ n_{ZnCl_2\left(2\right)}=n_{ZnO}=0,1\left(mol\right)\\ n_{ZnCl_2\left(tổng\right)}=0,2+0,1=0,3\left(mol\right)\\ m_{ddB}=m_{hhA}+m_{ddHCl}-m_{H_2}=21,1+200-0,2.2=220,7\left(g\right)\\ C\%_{ddZnCl_2}=\dfrac{136.0,3}{220,7}.100\%\approx18,487\%\)
\(n_{H_2}=\dfrac{4,48}{22,4}=0,2mol\\ Zn+2HCl\rightarrow ZnCl_2+H_2\left(1\right)\\ n_{Zn}=n_{H_2}=n_{ZnCl_2\left(1\right)}=0,2mol\\ n_{ZnO}=\dfrac{21,1-0,2.65}{81}=0,1mol\\ ZnO+2HCl\rightarrow ZnCl_2+H_2O\left(2\right)\\ n_{ZnCl_2\left(2\right)}=n_{ZnO}=0,1mol\\ C_{\%B}=C_{\%ZnCl_2}=\dfrac{\left(0,2+0,1\right).136}{21,1+200-0,2.2}\cdot100\%=18,49\%\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a.b.Hiện tượng: Kẽm tan dần trong dd, có chất khí thoát ra
\(n_{Zn}=\dfrac{6,5}{65}=0,1mol\)
\(Zn+2HCl\rightarrow ZnCl_2+H_2\)
0,1 0,2 0,1 ( mol )
\(V_{H_2}=0,1.22,4=2,24l\)
\(C_{M_{HCl}}=\dfrac{0,2}{0,3}=0,67M\)
c.\(CuO+H_2\rightarrow\left(t^o\right)Cu+H_2O\)
0,1 0,1 ( mol )
\(m_{Cu}=0,1.64=6,4g\)
`a)PTHH:`
`Zn + 2HCl -> ZnCl_2 + H_2 \uparrow`
`0,4` `0,8` `0,4` `0,4` `(mol)`
`n_[Zn]=26/65=0,4(mol)`
`b)m_[ZnCl_2]=0,4.136=54,4(g)`
`V_[H_2]=0,4.22,4=8,96(l)`
`c)m_[dd HCl]=[0,8.36,5]/10 . 100=292(g)`
`=>C%_[ZnCl_2]=[54,4]/[26+292-0,4.2] . 100~~17,15%`
`CuO` ở đâu vậy bạn