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\(a.Na_2CO_3+2HCl\rightarrow2NaCl+CO_2+H_2O\\ n_{NaCl}=n_{HCl}=2.n_{CO_2}=2.\dfrac{448:1000}{22,4}=0,04\left(mol\right)\\ C_{MddHCl}=\dfrac{0,04}{0,02}=2\left(M\right)\\ b.m_{NaCl}=58,5.0,04=2,34\left(g\right)\\ c.m_{Na_2CO_3}=106.0,02=2,12\left(g\right)\\ \%m_{Na_2CO_3}=\dfrac{2,12}{5}.100=42,4\%\\ \%m_{NaCl}=100\%-42,4\%=57,6\%\)
Bài 16 :
\(n_{CO2}=\dfrac{0,448}{22,4}=0,02\left(mol\right)\)
Pt : \(Na_2CO_3+2HCl\rightarrow2NaCl+CO_2+H_2O|\)
1 2 2 1 1
0,02 0,04 0,04 0,02
a) \(n_{HCl}=\dfrac{0,02.2}{1}=0,04\left(mol\right)\)
20ml = 0,02l
\(C_{M_{HCl}}=\dfrac{0,04}{0,02}=2\left(M\right)\)
b) \(n_{NaCl}=\dfrac{0,02.2}{1}=0,04\left(mol\right)\)
⇒ \(m_{NaCl}=0,04.58,5=2,34\left(g\right)\)
c) \(n_{Na2CO3}=\dfrac{0,04.1}{2}=0,02\left(mol\right)\)
⇒ \(m_{Na2CO3}=0,02.106=2,12\left(g\right)\)
\(m_{NaCl}=5-2,12=2,88\left(g\right)\)
0/0Na2CO3 = \(\dfrac{2,12.100}{5}=42,4\)0/0
0/0NaCl = \(\dfrac{2,88.100}{5}=57,6\)0/0
Chúc bạn học tốt
a, \(Zn+2HCl\rightarrow ZnCl_2+H_2\)
Ta có: \(n_{Zn}=\dfrac{13}{65}=0,2\left(mol\right)\)
Theo PT: \(n_{ZnCl_2}=n_{H_2}=n_{Zn}=0,2\left(mol\right)\)
\(\Rightarrow V_{H_2}=0,2.24,79=4,958\left(l\right)\)
b, \(n_{HCl}=2n_{Zn}=0,4\left(mol\right)\Rightarrow V_{HCl}=\dfrac{0,4}{2}=0,2\left(l\right)\)
c, \(C_{M_{ZnCl_2}}=\dfrac{0,2}{0,2}=1\left(M\right)\)
\(n_{Fe}=\dfrac{11,2}{56}=0,2\left(mol\right)\)
a.
\(Fe+2HCl\rightarrow FeCl_2+H_2\)
\(V_{H_2}=24,79.0,2=4,958\left(l\right)\)
b.
\(n_{HCl}=2.n_{Fe}=0,4\left(mol\right)\\ CM_{HCl}=\dfrac{0,4}{0,2}=2M\)
a, \(Zn+2HCl\rightarrow ZnCl_2+H_2\)
b, \(n_{Zn}=\dfrac{13}{65}=0,2\left(mol\right)\)
Theo PT: \(n_{HCl}=2n_{Zn}=0,4\left(mol\right)\Rightarrow C_{M_{HCl}}=\dfrac{0,4}{0,1}=4\left(M\right)\)
c, \(n_{H_2}=n_{Zn}=0,2\left(mol\right)\Rightarrow V_{H_2}=0,2.22,4=4,48\left(l\right)\)
PTHH: \(Fe+2HCl\rightarrow FeCl_2+H_2\uparrow\)
Ta có: \(n_{Fe}=\dfrac{5,6}{56}=0,1\left(mol\right)\)
\(\Rightarrow\left\{{}\begin{matrix}n_{HCl}=0,2\left(mol\right)\\n_{FeCl_2}=0,1\left(mol\right)=n_{H_2}\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}V_{H_2}=0,1\cdot22,4=2,24\left(l\right)\\m_{FeCl_2}=0,1\cdot127=12,7\left(g\right)\\C_{M_{FeCl_2}}=\dfrac{0,1}{0,1}=1\left(M\right)\\C_{M_{HCl}}=\dfrac{0,2}{0,1}=2\left(M\right)\end{matrix}\right.\)
Theo gt ta có: $n_{Al}=0,1(mol)$
a, $2Al+6HCl\rightarrow 2AlCl_3+3H_2$
b, $\Rightarrow n_{H_2}=0,15(mol)\Rightarrow V_{H_2}=3,36(l)$
c, Ta có: $n_{HCl}=0,3(mol)\Rightarrow m_{HCl}=10,95(g)\Rightarrow \%m_{ddHCl}=219(g)$
d, Bảo toàn khối lượng ta có: $m_{dd}=221,4(g)$
$\Rightarrow \%C_{AlCl_3}=6,02\%$
a.b.\(n_{Zn}=\dfrac{1,95}{65}=0,03mol\)
\(Zn+2HCl\rightarrow ZnCl_2+H_2\)
0,03 0,06 0,03 0,03 ( mol )
\(V_{H_2}=0,03.22,4=0,672l\)
\(m_{ddHCl}=\dfrac{0,06.36,5}{7,3\%}=30g\)
c.Tên muối: Kẽm clorua
\(m_{ZnCl_2}=0,03.136=4,08g\)
\(m_{ddspứ}=30+1,95-0,03.2=31,89g\)
\(C\%_{ZnCl_2}=\dfrac{4,08}{31,89}.100\%=12,79\%\)
\(a.PTHH:Zn+2HCl\xrightarrow[]{}ZnCl_2+H_2\\ b.n_{Zn}=\dfrac{13}{65}=0,2\left(mol\right)\\ PTHH:Zn+2HCl\xrightarrow[]{}ZnCl_2+H_2\\ n_{H_2}=0,2.2=0,4\left(mol\right)\\ V_{H_2}=0,4.22,4=8,96\left(l\right)\\ c.n_{HCl}=n_{Zn}=0,2mol\\ C_{MHCl}=\dfrac{0,4}{0,1}=4\left(M\right)\)
\(n_{Zn}=\dfrac{13}{65}=0,2\left(mol\right)\)
PTHH :
\(Zn+2HCl\rightarrow ZnCl_2+H_2\uparrow\)
0,2 0,4 0,2 0,2
\(b,V_{H_2}=0,2.22,4=4,48\left(l\right)\)
\(c,C_M=\dfrac{n}{V}=\dfrac{0,4}{0,1}=4M\)
\(n_{CaO}=\dfrac{11.2}{56}=0.2\left(mol\right)\)
\(CaO+2HCl\rightarrow CaCl_2+H_2O\)
\(0.2........0.4............0.2\)
\(n_{HCl}=0.4\left(mol\right)\)
\(m_{CaCl_2}=0.2\cdot111=22.2\left(g\right)\)
\(V_{dd_{HCl}}=\dfrac{0.4}{2}=0.2\left(l\right)\)
\(a)\ CaO + 2HCl \to CaCl_2 + H_2O\\ n_{CaO}= \dfrac{11,2}{56} = 0,2(mol)\\ n_{HCl} = 2n_{CaO} = 0,2.2 = 0,4(mol)\\ b)\ n_{CaCl_2} = n_{CaO} = 0,2(mol)\\ \Rightarrow m_{CaCl_2} = 0,2.111 = 22,2(gam)\\ c) V_{dd\ HCl} = \dfrac{0,4}{2} = 0,2(lít)\)