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\(Zn+2HCl\rightarrow ZnCl_2+H_2\uparrow\\ V_{HCl}=\dfrac{200}{1000}=0.2L\\ C_M=\dfrac{n_{ct}}{V_{HCl}}=\dfrac{\dfrac{6.5}{65}}{0.2}=0.5mol/l\\ n_{Zn}=\dfrac{m}{M}=\dfrac{6.5}{65}=0.1mol\rightarrow n_{H_2}=0.1mol\rightarrow V_{H_2}=n_{H_2}\cdot22.4=2.24L\)
200ml = 0,2l
\(Zn+2HCl\rightarrow ZnCl_2+H_2\uparrow\)
0,1.....0,2 0,1 (mol)
a) \(n_{Zn}=\dfrac{6,5}{65}=0,1\left(mol\right)\)
\(C_{M_{HCl}}=\dfrac{0,2}{0,2}=1\left(M\right)\)
b) \(V_{H_2\left(đktc\right)}=0,1.22,4=2,24\left(l\right)\)
\(V_{khí.thoát.ra}=V_{CH_4}=2,24l\)
\(n_{hh}=\dfrac{6,72}{22,4}=0,3mol\)
\(n_{CH_4}=\dfrac{2,24}{22,4}=0,1mol\)
\(\left\{{}\begin{matrix}\%V_{CH_4}=\dfrac{0,1}{0,3}.100=33,33\%\\\%V_{C_2H_4}=100\%-33,33\%=66,67\%\end{matrix}\right.\)
\(n_{C_2H_4}=0,3-0,1=0,2mol\)
\(200ml=0,2l\)
\(C_2H_4+Br_2\rightarrow C_2H_4Br_2\)
0,2 0,2 ( mol )
\(C_{MBr_2}=\dfrac{0,2}{0,2}=1M\)
\(n_{Fe}=\dfrac{84}{56}=1,5\left(mol\right)\\ PTHH:Fe+2HCl\rightarrow FeCl_2+H_2\\ n_{H_2}=n_{FeCl_2}=n_{Fe}=1,5\left(mol\right)\\ V_{H_2}=1,5.22,4=33,6\left(l\right)\\ C\%_{ddFeCl_2}=\dfrac{127.1,5}{84+300-1,5.2}.100\%=\dfrac{190,5}{381}.100\%=50\%\)
a) \(n_{Al}=\dfrac{5,4}{27}=0,2\left(mol\right)\)
PTHH: 2Al + 6HCl --> 2AlCl3 + 3H2
_____0,2-->0,6----->0,2---->0,3
=> \(C_{M\left(HCl\right)}=\dfrac{0,6}{0,2}=3M\)
b) VH2 = 0,3.24,79 = 7,437 (l)
c) \(C_{M\left(AlCl_3\right)}=\dfrac{0,2}{0,2}=1M\)
\(n_{HCl}=\dfrac{18.25}{36.5}=0.5\left(mol\right)\)
\(a.Mg+2HCl\rightarrow MgCl_2+H_2\)
\(b.\)
\(n_{Mg}=n_{H_2}=\dfrac{1}{2}\cdot n_{HCl}=\dfrac{1}{2}\cdot0.5=0.25\left(mol\right)\)
\(m_{Mg}=0.25\cdot24=6\left(g\right)\)
\(V_{H_2}=0.25\cdot22.4=5.6\left(l\right)\)
\(c.\)
\(V_{H_2\left(tt\right)}=5.6\cdot90\%=5.04\left(l\right)\)
\(a,PTHH:Na_2CO_3+2HCl\rightarrow2NaCl+H_2O+CO_2\uparrow\\ b,n_{Na_2CO_3}=\dfrac{15,9}{106}=0,15\left(mol\right)\\ \Rightarrow n_{HCl}=0,3\left(mol\right)\\ \Rightarrow m_{CT_{HCl}}=0,3\cdot36,5=10,95\left(g\right)\\ \Rightarrow C\%_{HCl}=\dfrac{10,95}{200}\cdot100\%=5,475\%\\ c,n_{CO_2}=0,15\left(mol\right)\\ \Rightarrow V_{CO_2\left(đkc\right)}=0,15\cdot24,79=3,7185\left(l\right)\\ d,m_{CO_2}=0,15\cdot44=6,6\left(g\right)\\ n_{NaCl}=0,3\left(mol\right);n_{H_2O}=0,15\left(mol\right)\\ \Rightarrow\left\{{}\begin{matrix}m_{CT_{NaCl}}=0,3\cdot58,5=17,55\left(g\right)\\m_{H_2O}=0,15\cdot18=2,7\left(g\right)\end{matrix}\right.\\ m_{dd_{NaCl}}=15,9+200-2,7-6,6=206,6\left(g\right)\\ \Rightarrow C\%_{NaCl}=\dfrac{17,55}{206,6}\cdot100\%\approx8,49\%\)
nFe=0,1 mol
Fe +2HCl=>FeCl2+H2
0,1 mol=>0,2 mol =>0,1 mol
VH2=0,1.22,4=2,24 lít
nHCl=0,2 mol=>mHCl=0,2.36,5=7,3g
=>C% dd HCl=7,3/200.100%=3,65%
a ,\(Zn+2HCl=>ZnCl_2+H_2\) (1)
b, \(n_{Zn}=\frac{6,5}{65}=0,1\left(mol\right)\)
theo (1) \(n_{H_2}=n_{Zn}=0,1\left(mol\right)\)
=> \(V_{H_2}=0,1.22,4=2,24\left(l\right)\)
c, Theo (1) \(n_{HCl}=2n_{Zn}=0,2\left(mol\right)\)
=> \(m_{HCl}=0,2.36,5=7,3 \left(g\right)\)
nồng độ % dung dịch axit đã dùng là
\(\frac{7,3}{200}.100\%=36,5\%\)