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a. \(n_{Zn}=\dfrac{6.5}{65}=0,1\left(mol\right)\)
PTHH : Zn + 2HCl -> ZnCl2 + H2
0,1 0,2 0,1
b. \(V_{H_2}=0,1.22,4=2,24\left(l\right)\)
c. \(m_{HCl}=0,2.36,5=7,3\left(g\right)\)
\(n_{Zn}=\dfrac{6,5}{65}=0,1mol\)
\(Zn+2HCl\rightarrow ZnCl_2+H_2\)
0,1 0,2 0,1
\(V_{H_2}=0,1\cdot22,4=2,24l\)
\(m_{HCl}=0,2\cdot36,5=7,3g\)
a) Pt: \(Fe+2HCl\rightarrow FeCl_2+H_2\)
b) nFe = \(\dfrac{11,2}{56}=0,2mol\)
Theo pt: nH2 = nFe = 0,2 mol
=> VH2 = 0,2.22,4 = 4,48lit
c) Theo pt: nHCl = 2nFe = 0,4 mol
=> mHCl = 0,4.36,5 = 14,6 g
=> C% = \(\dfrac{14,6}{73}.100\%=20\%\)
a) Pt: Fe+2HCl→FeCl2+H2Fe+2HCl→FeCl2+H2
b) nFe = \(\dfrac{11,2}{56}=0,2mol\)
Theo pt: nH2 = nFe = 0,2 mol
=> VH2 = 0,2.22,4 = 4,48lit
c) Theo pt: nHCl = 2nFe = 0,4 mol
=> mHCl = 0,4.36,5 = 14,6 g
=> \(C\%=\dfrac{14,6}{73}.100\%=20\%\)
\(1,\\ Mg+2HCl\rightarrow MgCl_2+H_2\\ 2,\\ n_{HCl}=\dfrac{18,25}{36,5}=0,5\left(mol\right)\Rightarrow n_{Mg}=n_{MgCl_2}=n_{H_2}=\dfrac{0,5}{2}=0,25\left(mol\right)\\ V_{H_2\left(đktc\right)}=0,25.22,4=5,6\left(l\right)\\ 3,\\ m_{MgCl_2}=95.0,25=23,75\left(g\right)\\ 4,\\ H_2+CuO\rightarrow\left(t^o\right)Cu+H_2O\\ n_{Cu}=n_{H_2}=0,25\left(g\right)\\ m_{Cu}=0,25.64=16\left(g\right)\)
1. \(Mg+2HCl\rightarrow MgCl_2+H_2\)
2. \(n_{HCl}=\dfrac{m_{HCl}}{M_{HCl}}=\dfrac{18,25}{36,5}\approx0,5\left(mol\right)\)
Theo PTHH: \(n_{H_2}=\dfrac{1}{2}n_{HCl}\)
\(\Rightarrow n_{H_2}=\dfrac{1}{2}.0,5=0,25\left(mol\right)\)
\(\Rightarrow V_{H_2}=n_{H_2}.22,4=0,25.22,4=5,6\left(l\right)\)
3. Theo PTHH: \(n_{MgCl_2}=\dfrac{1}{2}n_{HCl}\)
\(\Rightarrow n_{MgCl_2}=0,25\left(mol\right)\)
\(m_{MgCl_2}=n_{MgCl_2}.M_{MgCl_2}=0,25.95=23,75\left(g\right)\)
4. \(H_2+CuO\rightarrow Cu+H_2O\)
Theo PTHH: \(n_{Cu}=n_{H_2}=0,25\left(mol\right)\)
\(\Rightarrow m_{Cu}=n_{Cu}.M_{Cu}=0,25.64=16\left(g\right)\)
PTHH: \(Zn+2HCl\rightarrow ZnCl_2+H_2\uparrow\)
Ta có: \(n_{Zn}=\dfrac{0,65}{65}=0,01\left(mol\right)\)
\(\Rightarrow\left\{{}\begin{matrix}n_{HCl}=0,02\left(mol\right)\\n_{ZnCl_2}=0,01\left(mol\right)=n_{H_2}\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}C_{M_{HCl}}=\dfrac{0,02}{0,05}=0,4\left(M\right)\\m_{ZnCl_2}=0,01\cdot136=1,36\left(g\right)\\V_{H_2}=0,01\cdot22,4=0,224\left(l\right)\end{matrix}\right.\)
a,\(n_{Fe}=\dfrac{11,2}{56}=0,2\left(mol\right)\)
PTHH: Fe + 2HCl → FeCl2 + H2
Mol: 0,2 0,4 0,2
b, \(V_{H_2}=0,2.22,4=4,48\left(l\right)\)
c, \(m_{HCl}=0,4.36,5=14,6\left(g\right)\)
\(n_{Fe}=\dfrac{11.2}{56}=0.2\left(mol\right)\)
\(Fe+2HCl\rightarrow FeCl_2+H_2\)
\(0.2.......0.4....................0.2\)
\(V_{H_2}=0.2\cdot22.4=4.48\left(l\right)\)
\(m_{HCl}=0.4\cdot36.5=14.6\left(g\right)\)
PTHH : \(Fe+2HCl-->FeCl_2+H_2\uparrow\) (1)
\(n_{Fe}=\dfrac{m}{M}=\dfrac{11.2}{56}=0.2\left(mol\right)\)
Từ (1) => \(n_{Fe}=n_{H_2}=0.2\left(mol\right)\)
=> \(V_{H2\left(đktc\right)}=n.22,4=0,2.22,4=4,48\left(l\right)\)
Từ (1) => \(2n_{Fe}=n_{HCl}=0.4\left(mol\right)\)
=> \(m_{HCl}=n.M=0,4.\left(1+35.5\right)=14.6\left(g\right)\)
a)\(PTHH:Mg+2HCl->MgCl_2+H_2\)
b)\(n_{Mg}=\dfrac{1,2}{24}=0,05\left(m\right)\)
\(PTHH:Mg+2HCl->MgCl_2+H_2\)
tỉ lệ : 1 2 1 1
số mol:0,05 0,1 0,05 0,05
\(V_{H_2}=0,05.22,4=1,12\left(l\right)\)
c)\(m_{HCl}=0,1.36,5=3,65\left(g\right)\)