cho 195g Zn tác dụng hết với dung dịch HCl
a, viết ptpư
b, tính thể tích H2 thu được (đktc)
c, tính khối lượng Hcl cần dùng
MN GIÚP MÌNH VỚI MÌNH CẢM ƠN Ạ (ghi cụ thể đc ko ạ!)
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![](https://rs.olm.vn/images/avt/0.png?1311)
Zn+2HCl->ZnCl2+H2
0,3--0,6------0,3-----0,3
n Zn=\(\dfrac{19,5}{65}=0,3mol\)
=>VH2=0,3.22,4=6,72l
=>m HCl=0,6.36,5=21,9g
![](https://rs.olm.vn/images/avt/0.png?1311)
a+b+c) PTHH: \(Zn+2HCl\rightarrow ZnCl_2+H_2\uparrow\)
Ta có: \(n_{Zn}=\dfrac{6,5}{65}=0,1\left(mol\right)=n_{ZnCl_2}=n_{H_2}\)
\(\Rightarrow\left\{{}\begin{matrix}m_{ZnCl_2}=0,1\cdot136=13,6\left(g\right)\\V_{H_2}=0,1\cdot22,4=2,24\left(l\right)\end{matrix}\right.\)
d) PTHH: \(CuO+H_2\underrightarrow{t^o}Cu+H_2O\)
Theo PTHH: \(n_{H_2}=n_{Cu}=0,1\left(mol\right)\)
\(\Rightarrow m_{Cu}=0,1\cdot64=6,4\left(g\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a, PT: \(Zn+2HCl\rightarrow ZnCl_2+H_2\)
b, Ta có: \(n_{Zn}=\dfrac{6,5}{65}=0,1\left(mol\right)\)
Theo PT: \(n_{H_2}=n_{Zn}=0,1\left(mol\right)\Rightarrow V_{H_2}=0,1.22,4=2,24\left(l\right)\)
c, PT: \(Fe_2O_3+3H_2\underrightarrow{t^o}2Fe+3H_2O\)
Theo PT: \(n_{Fe}=\dfrac{2}{3}n_{H_2}=\dfrac{1}{15}\left(mol\right)\Rightarrow m_{Fe}=\dfrac{1}{15}.56=\dfrac{56}{15}\left(g\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(Fe+2HCl\text{→}FeCl_2+H_2\)
n Fe = 2,8:56=0,05 mol = n H2
V H2 = 0,05.22,4=1,12 lít
n HCl = n Fe .2 =0,1 mol
m HCl = 0,1.(1+35,5)=3,65 g
a) PTHH: \(Fe+2HCl\rightarrow FeCl_2+H_2\uparrow\)
b) \(n_{H_2}=n_{Fe}=\dfrac{2,8}{56}=0,05\left(mol\right)\)
\(\Rightarrow V_{H_2}=n.22,4=0,05.22,4=1,12\left(l\right)\)
c) \(n_{HCl}=2n_{Fe}=0,1\left(mol\right)\)
\(m_{HCl}=n.M=0,1.36,5=3,65\left(g\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(n_{Al}=\dfrac{7,5.36\%}{27}=0,1\left(mol\right)\)
\(n_{Mg}=\dfrac{7,5-0,1.27}{24}=0,2\left(mol\right)\)
PTHH: 2Al + 6HCl --> 2AlCl3 + 3H2
0,1------------>0,1----->0,15
Mg + 2HCl --> MgCl2 + H2
0,2------------>0,2----->0,2
=> mmuối = 0,1.133,5 + 0,2.95 = 32,35 (g)
b) VH2 = (0,15 + 0,2).22,4 = 7,84 (l)
![](https://rs.olm.vn/images/avt/0.png?1311)
a, \(n_{Zn}=\dfrac{19,5}{65}=0,3\left(mol\right)\)
PT: \(Zn+2HCl\rightarrow ZnCl_2+H_2\)
Theo PT: \(n_{ZnCl_2}=n_{Zn}=0,3\left(mol\right)\Rightarrow m_{ZnCl_2}=0,3.136=40,8\left(g\right)\)
b, \(n_{H_2}=n_{Zn}=0,3\left(mol\right)\Rightarrow V_{H_2}=0,3.22,4=6,72\left(l\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) $Zn + 2HCl \to ZnCl_2 + H_2$
b) Theo PTHH :
$n_{ZnCl_2} = n_{Zn} = \dfrac{6,5}{65} = 0,1(mol)$
$m_{ZnCl_2} = 0,1.136 = 13,6(gam)$
c)
$n_{H_2} = n_{Zn} = 0,1(mol) \Rightarrow V_{H_2} = 0,1.22,4 = 2,24(lít)$
d)
$n_{HCl} = 2n_{Zn} = 0,2(mol) \Rightarrow C_{M_{HCl}} = \dfrac{0,2}{0,5} = 0,4M$
![](https://rs.olm.vn/images/avt/0.png?1311)
PTHH: \(Zn+2HCl\rightarrow ZnCl_2+H_2\uparrow\)
Ta có: \(n_{Zn}=\dfrac{32,5}{65}=0,5\left(mol\right)\)
\(\Rightarrow\left\{{}\begin{matrix}n_{HCl}=1\left(mol\right)\\n_{H_2}=0,5\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}m_{ddHCl}=\dfrac{1\cdot36,5}{20\%}=182,5\left(g\right)\\V_{H_2}=0,5\cdot22,4=11,2\left(l\right)\end{matrix}\right.\)
Zn+2Hcl->Zncl2+H2
3----6--------3--------3
n Zn=\(\dfrac{195}{65}\)=3 mol
=>VH2=3.22,4=67,2l
=>mHCl=3.36,5=109,5g
Zn+2Hcl->Zncl2+H2
3-----6------3------ 3
nZn=195/65=3 (mol)
=>VH2=3.22,4=67,2 (l)
=>mHCl=3.36,5=109,5 (g)