\(n_{Na}=\dfrac{9,2}{23}=0,4mol\)

\(Na+CH_3COOH\rightarrow CH_3COONa+\dfrac{1}{2}H_2\)

0,4             0,4                                        ( mol )

\(m_{CH_3COOH}=0,4.60=24g\)

$a\big)$

$n_{Na}=\dfrac{4,6}{23}=0,2(mol)$

$CH_3COOH+Na\to CH_3COONa+\dfrac{1}{2}H_2$

Theo PT: $n_{CH_3COOH}=n_{Na}=0,2(mol)$

$\to m_{CH_3COOH}=0,2.60=12(g)$

$b\big)$

Theo PT: $n_{H_2}=\dfrac{1}{2}n_{Na}=0,1(mol)$

$\to V_{H_2(đktc)}=0,1.22,4=2,24(l)$

giúp tuiii

\(Na+H_2O\rightarrow NaOH+\dfrac{1}{2}H_2\)

Bảo toàn khối lượng : 

\(m_{Na}+m_{H_2O}=m_{NaOH}+m_{H_2}\)

\(m_{NaOH}=2.3+100-0.1=102.2\left(g\right)\)

\(n_{CO_2\left(đktc\right)}=\dfrac{10,64}{22,4}=0,475\left(mol\right)\\a, K_2CO_3+2CH_3COOH\rightarrow2CH_3COOK+CO_2+H_2O\\ b,n_{K_2CO_3}=n_{CO_2}=0,475\left(mol\right)\\ \Rightarrow m_{K_2CO_3}=138.0,475=65,55\left(g\right)\\ n_{CH_3COOH}=0,475.2=0,95\left(mol\right)\\ C\%_{ddCH_3COOH}=\dfrac{0,95.60}{200}.100=28,5\%\)

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!

\(n_{Mg}=\dfrac{4,8}{24}=0,2\left(mol\right)\\ a,PTHH:Mg+2CH_3COOH\rightarrow\left(CH_3COO\right)_2Mg+H_2\\ n_{H_2}=n_{\left(CH_3COO\right)_2Mg}=n_{Mg}=0,2\left(mol\right);n_{CH_3COOH}=2.0,2=0,4\left(mol\right)\\ b,V_{H_2\left(đktc\right)}=0,2.22,4=4,48\left(l\right)\\ c,m_{\left(CH_3COO\right)_2Mg}=142.0,2=28,4\left(g\right)\\ d,m_{ddCH_3COOH}=\dfrac{0,4.60.100}{12}=200\left(g\right)\)

CH3CHO + 2AgNO3 + 3NH3 + H2O → CH3COONH4 + 2Ag↓ + 2NH4NO3    (1)

CH3COOH + NaOH → CH3COONa + H2O    (2)