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Phương trình hóa học:
2CH3COOH + K2CO3 => CH3COOK + CO2 + H2O
nCO2 = V/22.4 = 10.64/22.4 = 0.475 (mol)
Theo phương trình => nCH3COOH = 0.95 (mol)
a = mCH3COOH = n.M = 0.95 x 60 = 57 (g)
C% = 57 x 100/200 = 28.5 (%)
$a\big)$
$Zn+2CH_3COOH\to (CH_3COO)_2Zn+H_2$
$ZnO+2CH_3COOH\to (CH_2COO)_2Zn+H_2O$
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 )
\(C_{M\left(CH_3COOH\right)}=\dfrac{0,4+0,2}{0,2}=3M\)
\(n_{CH_3COOH}=\dfrac{130.12}{100.60}=0,26\left(mol\right)\)
\(CaCO_3+2CH_3COOH\rightarrow\left(CH_3COO\right)_2Ca+CO_2+H_2O\)
0,13 0,26 0,13 0,13 ( mol )
\(m_{CaCO_3}=0,13.100=13\left(g\right)\)
\(V_{CO_2}=0,13.22,4=2,912\left(l\right)\)
\(m_{ddspứ}=13+130-0,13.44=137,28\left(g\right)\)
\(C\%_{\left(CH_3COO\right)_3Ca}=\dfrac{0,13.158}{137,28}.100=14,96\%\)
\(n_{Na}=\dfrac{9,2}{23}=0,4mol\)
\(Na+CH_3COOH\rightarrow CH_3COONa+\dfrac{1}{2}H_2\)
0,4 0,4 ( mol )
\(m_{CH_3COOH}=0,4.60=24g\)
a)
$Fe + H_2SO_4 \to FeSO_4 + H_2$
$n_{H_2SO_4} = n_{H_2} = n_{Fe} = \dfrac{16,8}{56} = 0,3(mol)$
$V = 0,3.22,4 = 6,72(lít)$
$C_{M_{H_2SO_4}} = \dfrac{0,3}{0,25} = 1,2M$
b)
$n_{CuO} = \dfrac{16}{80} = 0,2(mol)$
$CuO + H_2 \xrightarrow{t^o} Cu + H_2O$
$n_{CuO} < n_{H_2}$ nên $H_2$ dư
$n_{Cu} = n_{CuO} = 0,2(mol)$
$m_{Cu} = 0,2.64 = 12,8(gam)$
\(n_{Fe}=\dfrac{m}{M}=\dfrac{16,8}{56}=0,3\left(mol\right)\\ PT:Fe+H_2SO_4\rightarrow FeSO_4+H_2\)
0,3 0,3 0,3 (mol)
a) V= n. 22,4 = 0,3 . 22,4 = 6,72(l)
\(C\%=\dfrac{m_{H_2SO_4}}{m_{dd}}.100\%=\dfrac{0,3.98}{200}.100\%=11,76\%\)
b) PT: \(H_2+CuO\underrightarrow{t^o}Cu+H_2O\)
0,3 0,3
=> mCu=n.M=0,3.64=19,2(g)
\(Fe+2CH_3COOH\rightarrow\left(CH_3COO\right)_2Fe+H_2\)
\(n_{Fe}=\dfrac{11,2}{56}=0,2\left(mol\right)\)
Theo PT: \(n_{\left(CH_3COO\right)_2Fe}=n_{Fe}=0,2\left(mol\right)\Rightarrow m_{\left(CH_3COO\right)_2Fe}=0,2.174=34,8\left(g\right)\)
Ta có: \(n_{H_2}=n_{Fe}=0,2\left(mol\right)\)
m dd sau pư = 11,2 + 200 - 0,2.2 = 210,8 (g)
\(\Rightarrow C\%_{\left(CH_3COO\right)_2Fe}=\dfrac{34,8}{210,8}.100\%\approx16,51\%\)
\(n_{CO_2\left(đktc\right)}=\dfrac{10,64}{22,4}=0,475\left(mol\right)\\a, K_2CO_3+2CH_3COOH\rightarrow2CH_3COOK+CO_2+H_2O\\ b,n_{K_2CO_3}=n_{CO_2}=0,475\left(mol\right)\\ \Rightarrow m_{K_2CO_3}=138.0,475=65,55\left(g\right)\\ n_{CH_3COOH}=0,475.2=0,95\left(mol\right)\\ C\%_{ddCH_3COOH}=\dfrac{0,95.60}{200}.100=28,5\%\)