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\(n_{Zn}=\dfrac{13}{65}=0,2\left(mol\right)\)
a) PTHH : \(Zn+2HCl\rightarrow ZnCl_2+H_2\)
b) \(n_{HCl}=2n_{Zn}=0,4\left(mol\right)\Rightarrow C_{MddHCl}=\dfrac{0,4}{0,2}=2M\)
c) \(n_{H2}=n_{Zn}=0,2\left(mol\right)\Rightarrow V_{H2\left(dkc\right)}=0,2.24,79=4,958\left(l\right)\)
\(a,PTHH:Zn+2HCl\rightarrow ZnCl_2+H_2\\ b,n_{Zn}=\dfrac{13}{65}=0,2\left(mol\right)\\ n_{HCl}=2\cdot0,15=0,3\left(mol\right)\)
Vì \(\dfrac{n_{Zn}}{1}>\dfrac{n_{HCl}}{2}\) nên sau p/ứ Zn dư
\(\Rightarrow n_{Zn}=\dfrac{1}{2}n_{HCl}=0,15\left(mol\right)\\ \Rightarrow m_{Zn}=0,15\cdot65=9,75\\ \Rightarrow m_{Zn\left(dư\right)}=13-9,75=3,25\left(g\right)\\ c,n_{H_2}=n_{Zn}=0,15\left(mol\right)\\ \Rightarrow V_{H_2}=0,15\cdot22,4=3,36\left(l\right)\)
\(n_{K2O}=\dfrac{9,4}{94}=0,1\left(mol\right)\)
Pt : \(K_2O+H_2O\rightarrow2KOH|\)
1 1 2
0,1 0,2
a) \(n_{KOH}=\dfrac{0,1.2}{1}=0,2\left(mol\right)\)
200ml = 0,2l
\(C_{M_{ddKOH}}=\dfrac{0,2}{0,2}=1\left(M\right)\)
b) Pt : \(KOH+HCl\rightarrow KCl+H_2O|\)
1 1 1 1
0,1 0,1
\(n_{HCl}=\dfrac{0,1.1}{1}=0,1\left(mol\right)\)
\(m_{HCl}=0,1.36,5=3,65\left(g\right)\)
\(m_{ddHCl}=\dfrac{3,65.100}{10}=36,5\left(g\right)\)
c) \(CO_2+2KOH\rightarrow K_2CO_3+H_2O|\)
1 2 1 1
0,05 0,1
\(n_{CO2}=\dfrac{0,1.1}{2}=0,05\left(mol\right)\)
\(V_{CO2\left(dktc\right)}=0,05.22,4=1,12\left(l\right)\)
Chúc bạn học tốt
a, \(Zn+2HCl\rightarrow ZnCl_2+H_2\)
b, \(n_{H_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\)
Theo PT: \(n_{Zn}=n_{H_2}=0,3\left(mol\right)\)
\(\Rightarrow m_{Zn}=0,3.65=19,5\left(g\right)\)
c, \(n_{HCl}=2n_{H_2}=0,6\left(mol\right)\)
\(\Rightarrow C_{M_{HCl}}=\dfrac{0,6}{0,1}=6\left(M\right)\)
1 , \(n_{Na}=\frac{4,6}{23}=0,2\left(mol\right)\)
\(m_{HCl}=200.2,92\%=5,84\left(mol\right)\) => \(n_{HCl}=\frac{5,84}{36,5}=0,16\left(mol\right)\)
\(2Na+2HCl->2NaCl+H_2\left(1\right)\)
vì \(\frac{0,2}{2}>\frac{0,16}{2}\) => Na dư , HCl hết
dung dịch thu được là dung dịch NaCl
theo (1) \(n_{NaCl}=n_{HCl}=0,16\left(mol\right)\) => \(m_{NaCl}=0,16.58,5=9,36\left(g\right)\)
\(n_{H_2}=\frac{1}{2}n_{HCl}=0,08\left(mol\right)\)
khối lượng dung dịch sau phản ứng là
4,6+200-0,08.2=204,44(g)
\(C_{\%\left(NaCl\right)}=\frac{9,36}{204,44}.100\%\approx4,58\%\)
$a\big)$
$Zn+2CH_3COOH\to (CH_3COO)_2Zn+H_2$
$ZnO+2CH_3COOH\to (CH_2COO)_2Zn+H_2O$
Theo PT: $n_{Zn}=n_{H_2}=\frac{4,48}{22,4}=0,2(mol)$
$\to \%m_{Zn}=\frac{0,2.65}{21,1}.100\%\approx 61,61\%$
$\to \%m_{ZnO}=100-61,61=38,39\%$
$b\big)$
$n_{ZnO}=\frac{21,1-0,2.65}{81}=0,1(mol)$
Theo PT: $\sum n_{CH_3COOH}=2n_{Zn}+2n_{ZnO}=0,6(mol)$
$\to C_{M_{CH_3COOH}}=\dfrac{0,6}{\frac{200}{1000}}=3M$
\(n_{H_2}=\dfrac{4,48}{22,4}=0,2mol\)
\(Zn+2CH_3COOH\rightarrow\left(CH_3COO\right)_2Zn+H_2\)
0,2 0,2 ( mol )
\(m_{Zn}=0,2.65=13g\)
\(\%m_{Zn}=\dfrac{13}{21,1}.100=61,61\%\)
\(\%m_{ZnO}=100\%-61,61\%=38,39\%\)
\(Zn+2CH_3COOH\rightarrow\left(CH_3COO\right)_2Zn+H_2\)
0,2 0,4 ( mol )
\(n_{ZnO}=\dfrac{21,1-13}{81}=0,1mol\)
\(ZnO+2CH_3COOH\rightarrow\left(CH_3COO\right)_2Zn+H_2O\)
0,1 0,2 ( mol )
\(C_{M\left(CH_3COOH\right)}=\dfrac{0,4+0,2}{0,2}=3M\)
a)
$CaCO_3 + 2HCl \to CaCl_2 + CO_2 + H_2O$
$n_{CaCO_3} = n_{CO_2} = \dfrac{672}{1000.22,4} = 0,03(mol)$
$n_{HCl} = 2n_{CO_2} = 0,06(mol)$
$\Rightarrow C_{M_{HCl}} = \dfrac{0,06}{0,2} = 0,3M$
b)
$\%m_{CaCO_3} = \dfrac{0,03.100}{5}.100\% = 60\%$
$\%m_{CaSO_4}= 100\% -60\% = 40\%$
\(n_{Zn}=\dfrac{2,6}{65}=0,04\left(mol\right)\)
Pt : \(Zn+2HCl\rightarrow ZnCl_2+H_2\uparrow\)
a) Sửa đề : tính khối lượng ZnCl2 thu được
Theo Pt : \(n_{Zn}=n_{ZnCl2}=n_{H2}=0,04\left(mol\right)\)
\(\Rightarrow m_{ZnCl2}=0,04.136=5,44\left(g\right)\)
b) Theo Pt : \(n_{HCl}=2n_{Zn}=2.0,04=0,08\left(mol\right)\)\
\(C_{MddHCl}=\dfrac{0,08}{0,2}=0,4\left(M\right)\)
c) \(V_{H2\left(dktc\right)}=0,04.22,4=0,896\left(l\right)\)
Chúc bạn học tốt
thank bạn nhiều ạ