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$a\big)$
$n_{Na}=\dfrac{4,6}{23}=0,2(mol)$
$CH_3COOH+Na\to CH_3COONa+\dfrac{1}{2}H_2$
Theo PT: $n_{CH_3COOH}=n_{Na}=0,2(mol)$
$\to m_{CH_3COOH}=0,2.60=12(g)$
$b\big)$
Theo PT: $n_{H_2}=\dfrac{1}{2}n_{Na}=0,1(mol)$
$\to V_{H_2(đktc)}=0,1.22,4=2,24(l)$
\(n_{CO_2\left(đktc\right)}=\dfrac{10,64}{22,4}=0,475\left(mol\right)\\a, K_2CO_3+2CH_3COOH\rightarrow2CH_3COOK+CO_2+H_2O\\ b,n_{K_2CO_3}=n_{CO_2}=0,475\left(mol\right)\\ \Rightarrow m_{K_2CO_3}=138.0,475=65,55\left(g\right)\\ n_{CH_3COOH}=0,475.2=0,95\left(mol\right)\\ C\%_{ddCH_3COOH}=\dfrac{0,95.60}{200}.100=28,5\%\)
a) PTHH: \(Mg+2HCl\rightarrow MgCl_2+H_2\uparrow\)
b) Ta có: \(n_{Mg}=\dfrac{3,6}{24}=0,15\left(mol\right)=n_{MgCl_2}\)
\(\Rightarrow m_{MgCl_2}=0,15\cdot95=14,25\left(g\right)\)
c) Theo PTHH: \(\left\{{}\begin{matrix}n_{HCl\left(p.ứ\right)}=0,3\left(mol\right)\\n_{H_2}=0,15\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}m_{HCl\left(p.ứ\right)}=0,3\cdot36,5=10,95\left(g\right)\\m_{H_2}=0,15\cdot2=0,3\left(g\right)\end{matrix}\right.\)
Mặt khác: \(m_{dd}=m_{Mg}+m_{ddHCl}-m_{H_2}=53,3\left(g\right)\)
\(\Rightarrow m_{ddHCl}=50\left(g\right)\) \(\Rightarrow C\%_{HCl\left(p.ứ\right)}=\dfrac{10,95}{50}\cdot100\%=21,9\%\)
\(n_{Zn}=\dfrac{19,5}{65}=0,3\left(mol\right)\\ a,PTHH:Zn+2CH_3COOH\rightarrow\left(CH_3COO\right)_2Zn+H_2\\ b,n_{H_2}=n_{\left(CH_3COO\right)_2Zn}=n_{Zn}=0,3\left(mol\right);n_{CH_3COOH}=2.0,3=0,6\left(mol\right)\\ b,V_{H_2\left(đkc\right)}=24,79.0,3=7,437\left(l\right)\\ c,m_{\left(CH_3COO\right)_2Zn}=0,3.183=54,9\left(g\right)\\ d,C_2H_5OH+CH_3COOH\rightarrow CH_3COOC_2H_5+H_2O\\ n_{Este\left(LT\right)}=n_{CH_3COOH}=0,6\left(mol\right)\\ n_{este\left(TT\right)}=80\%.0,6=0,48\left(mol\right)\\ m=m_{este\left(TT\right)}=88.0,48=42,24\left(g\right)\)
Ta có: \(n_K=\dfrac{7,8}{39}=0,2\left(mol\right)\)
a, PT: \(2K+2CH_3COOH\rightarrow2CH_3COOK+H_2\)
_____0,2______0,2_____________________0,1 (mol)
b, \(m_{CH_3COOH}=0,2.60=12\left(g\right)\)
c, \(V_{H_2}=0,1.22,4=2,24\left(l\right)\)
Bạn tham khảo nhé!
\(n_{Mg}=\dfrac{4,8}{24}=0,2\left(mol\right)\\ a,PTHH:Mg+2CH_3COOH\rightarrow\left(CH_3COO\right)_2Mg+H_2\\ n_{H_2}=n_{\left(CH_3COO\right)_2Mg}=n_{Mg}=0,2\left(mol\right);n_{CH_3COOH}=2.0,2=0,4\left(mol\right)\\ b,V_{H_2\left(đktc\right)}=0,2.22,4=4,48\left(l\right)\\ c,m_{\left(CH_3COO\right)_2Mg}=142.0,2=28,4\left(g\right)\\ d,m_{ddCH_3COOH}=\dfrac{0,4.60.100}{12}=200\left(g\right)\)
a, \(Zn+2CH_3COOH\rightarrow\left(CH_3COO\right)_2Zn+H_2\)
\(ZnO+2CH_3COOH\rightarrow\left(CH_3COO\right)_2Zn+H_2O\)
b, Ta có: \(n_{H_2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\)
Theo PT: \(n_{Zn}=n_{H_2}=0,1\left(mol\right)\)
\(\Rightarrow\left\{{}\begin{matrix}\%m_{Zn}=\dfrac{0,1.65}{14,6}.100\%\approx44,52\%\\\%m_{ZnO}\approx55,48\%\end{matrix}\right.\)
c, Ta có: \(n_{ZnO}=\dfrac{14,6-0,1.65}{81}=0,1\left(mol\right)\)
Theo PT: \(n_{CH_3COOH}=2n_{Zn}+2n_{ZnO}=0,4\left(mol\right)\)
PT: \(C_2H_5OH+O_2\underrightarrow{^{mengiam}}CH_3COOH+H_2O\)
Theo PT: \(n_{C_2H_5OH\left(LT\right)}=n_{CH_3COOH}=0,4\left(mol\right)\)
Mà: H = 80%
\(\Rightarrow n_{C_2H_5OH\left(TT\right)}=\dfrac{0,4}{80\%}=0,5\left(mol\right)\)
\(\Rightarrow m_{C_2H_5OH\left(TT\right)}=0,5.46=23\left(g\right)\)
\(\Rightarrow V_{C_2H_5OH}=\dfrac{23}{0,8}=28,75\left(ml\right)\)
\(\Rightarrow V_{C_2H_5OH\left(18,4^o\right)}=\dfrac{28,75}{18,4}.100=156,25\left(ml\right)=0,15625\left(l\right)\)
\(n_{Na}=\dfrac{9,2}{23}=0,4mol\)
\(Na+CH_3COOH\rightarrow CH_3COONa+\dfrac{1}{2}H_2\)
0,4 0,4 ( mol )
\(m_{CH_3COOH}=0,4.60=24g\)