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\(a) n_{Zn} = \dfrac{19,5}{65} = 0,3(mol) ; n_{HCl} = 0,35.2 = 0,7(mol)\\ Zn + 2HCl \to ZnCl_2 + H_2\\ n_{HCl} = 0,7 > 2n_{Zn} = 0,6 \to HCl\ dư\\ n_{H_2} = n_{Zn} = 0,3(mol) \Rightarrow V_{H_2} = 0,3.22,4 = 6,72(lít)\\ b) m_{dd\ HCl} = 350.1,05 = 367,5(gam)\\ m_{dd\ sau\ pư} = 19,5 + 367,5 - 0,3.2 = 386,4(gam)\\ \Rightarrow C\%_{ZnCl_2} = \dfrac{0,3.136}{386,4}.100\% = 10,56\%\\ c) C\%_{HCl} = \dfrac{0,7.36,5}{367,5}.100\% = 6,95\%\)
\(n_{Fe} = \dfrac{2,8}{56} = 0,05(mol)\\ Fe + 2HCl \to FeCl_2 + H_2\\ n_{HCl} =2n_{Fe} = 0,05.2 = 0,1(mol)\\ V_{dd\ HCl} = \dfrac{0,1}{2} = 0,05(lít)\)
a, \(n_{Mg}=\dfrac{2,4}{24}=0,1\left(mol\right)\)
PT: \(Mg+2HCl\rightarrow MgCl_2+H_2\)
Theo PT: \(n_{H_2}=n_{Mg}=0,1\left(mol\right)\Rightarrow V_{H_2}=0,1.22,4=2,24\left(l\right)\)
b, \(n_{HCl}=2n_{Mg}=0,2\left(mol\right)\Rightarrow V_{HCl}=\dfrac{0,2}{2}=0,1\left(l\right)\)
\(n_{Mg}=\dfrac{2,4}{24}=0,1\left(mol\right)\)
PTHH :
\(Mg+2HCl\rightarrow MgCl_2+H_2\uparrow\)
0,1 0,2 0,1
\(a,V_{H_2}=0,1.22,4=2,24\left(l\right)\)
\(b,V=\dfrac{n}{C_M}=\dfrac{0,2}{2}=0,1\left(l\right)\)
\(a) Fe + 2HCl \to FeCl_2 + H_2\\ n_{FeCl_2} = n_{H_2} = n_{Fe} = \dfrac{5,6}{56} = 0,1(mol)\\ V_{H_2} = 0,1.22,4 = 2,24(lít)\\ b)m_{FeCl_2} = 0,1.127 = 12,7(gam)\\ c) n_{HCl} =2 n_{Fe} = 0,2(mol)\\ C_{M_{HCl}} = \dfrac{0,2}{0,2} = 1M\)
nZn = 6.5/65 = 0.1 (mol)
Zn + 2HCl => ZnCl2 + H2
0.1.......0.2...................0.1
VddHCl = 0.2/2 = 0.1 (l)
nFe = 3/56 (mol)
Fe2O3 + 3H2 -to-> 2Fe + 3H2O
.................9/112........3/56
H% = 9/112 / 0.1 * 100% = 80.35%
a) Zn + 2HCl $\to$ ZnCl2 + H2
b) n Zn = 6,5/65 = 0,1(mol)
Theo PTHH : n HCl = 2n Zn = 0,2(mol)
=> V dd HCl = 0,2/2 = 0,1(lít)
c) n Fe = 3/56 (mol)
Fe2O3 + 3H2 $\xrightarrow{t^o}$ 2Fe + 3H2O
Theo PTHH :
n H2 = 3/2 n Fe = 9/112(mol)
Vậy :
H = $\dfrac{ \dfrac{9}{112} }{0,1}$ .100% = 80,36%
1/
\(CaCO_3 + 2HCl \to CaCl_2 + CO_2 +H_2O\\ CO_2 + NaOH \to NaHCO_3 2NaHCO_3 \xrightarrow{t^o} Na_2CO_3 + CO_2 + H_2O\\ Na_2CO_3 + BaCl_2 \to BaCO_3 + 2NaCl\)
1)
\(CaCO_3\underrightarrow{t^o}CO_2+CaO\\ NaOH+CO_2\rightarrow NaHCO_3\\ NaHCO_3+NaOH\rightarrow Na_2CO_3+H_2O\\ Na_2CO_3+Ba\left(OH\right)_2\rightarrow NaOH+BaCO_3\)
2)
\(n_{HCl}=C_{M_{HCl}}.V_{HCl}=1.0,2=0,2\left(mol\right)\)
PTHH: \(K_2CO_3+2HCl\rightarrow2KCl+H_2O+CO_2\)
\(\Rightarrow n_{K_2CO_3}=n_{CO_2}=0,1\left(mol\right)\)
a) \(V_{CO_2\left(đktc\right)}=0,1.22,4=2,24\left(l\right)\)
b) \(m_{K_2CO_3}=0,1.138=13,8\left(g\right)\)
\(m_{ddK_2CO_3}=\dfrac{13,8.100}{13,8}=100\left(g\right)\)
a, \(n_{Al}=\dfrac{5,4}{27}=0,2\left(mol\right)\)
PTHH: 2Al + 6HCl → 2AlCl3 + 3H2
Mol: 0,2 0,6 0,2 0,3
\(m_{AlCl_3}=0,2.133,5=26,7\left(g\right)\)
b, \(V_{H_2}=0,3.22,4=6,72\left(l\right)\)
c, \(m_{ddHCl}=\dfrac{0,6.36,5.100}{10}=219\left(g\right)\)
a,\(n_{Fe}=\dfrac{28}{56}=0,5\left(mol\right)\)
PTHH: Fe + 2HCl → FeCl2 + H2
Mol: 0,5 1 0,5 0,5
\(V_{ddHCl}=\dfrac{1}{2}=0,5\left(l\right)\)
b,\(V_{H_2}=0,5.22,4=11,2\left(l\right)\)
c,\(C_{M_{ddFeCl_2}}=\dfrac{0,5}{0,5}=1M\)
