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\(n_{CH_3COOH}=0.5\cdot1=0.5\left(mol\right)\)
\(CH_3COOH+NaOH\rightarrow CH_3COONa+H_2O\)
\(0.5.....................0.5..................0.5\)
\(m_{CH_3COOH}=0.5\cdot60=30\left(g\right)\)
\(m_{CH_3COONa}=0.5\cdot82=41\left(g\right)\)
\(m_{NaOH}=0.5\cdot40=20\left(g\right)\)
\(n_{Na_2CO_3}=0.1\cdot0.5=0.05\left(mol\right)\)
\(2CH_3COOH+Na_2CO_3\rightarrow2CH_3COONa+CO_2+H_2O\)
\(2.........................1\)
\(0.5.......................0.05\)
\(LTL:\dfrac{0.5}{2}>\dfrac{0.05}{1}\Rightarrow CH_3COOHdư\)
\(V_{CO_2}=0.05\cdot22.4=1.12\left(l\right)\)
a, \(n_{CO_2}=\dfrac{2,479}{24,79}=0,1\left(mol\right)\)
PT: \(MgO+2CH_3COOH\rightarrow\left(CH_3COO\right)_2Mg+H_2O\)
\(MgCO_3+2CH_3COOH\rightarrow\left(CH_3COO\right)_2Mg+CO_2+H_2O\)
Theo PT: \(n_{MgCO_3}=n_{CO_2}=0,1\left(mol\right)\)
\(\Rightarrow\left\{{}\begin{matrix}\%m_{MgCO_3}=\dfrac{0,1.84}{10,4}.100\%\approx80,77\%\\\%m_{MgO}\approx19,23\%\end{matrix}\right.\)
b, \(n_{MgO}=\dfrac{10,4-0,1.84}{40}=0,05\left(mol\right)\)
Theo PT: \(n_{CH_3COOH}=2n_{MgO}+2n_{MgCO_3}=0,3\left(mol\right)\)
\(\Rightarrow C_{M_{CH_3COOH}}=\dfrac{0,3}{0,2}=1,5\left(M\right)\)
Ta có: \(n_{NaOH}=0,1.0,5=0,05\left(mol\right)\)
PT: \(CH_3COOH+NaOH\rightarrow CH_3COONa+H_2O\)
Theo PT: \(n_{CH_3COOH}=n_{CH_3COONa}=n_{NaOH}=0,05\left(mol\right)\)
a, \(C_{M_{CH_3COOH}}=\dfrac{0,05}{0,2}=0,25\left(M\right)\)
b, \(m_{CH_3COONa}=0,05.82=4,1\left(g\right)\)
\(n_{Na_2CO_3}=\dfrac{21,6}{106}\)
Số lẻ lắm em, em xem 21,6 hay 21,2 gam nhé!
a)
\(n_{CO_2}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\)
PTHH: Na2CO3 + 2CH3COOH --> 2CH3COONa + CO2 + H2O
0,15<---------0,3<------------------------0,15
=> \(C\%_{dd.CH_3COOH}=\dfrac{0,3.60}{200}.100\%=9\%\)
b)
\(m_{dd.Na_2CO_3}=\dfrac{0,15.106.100}{15}=106\left(g\right)\)
c)
PTHH: 2CH3COOH + Ba(OH)2 --> (CH3COO)2Ba + 2H2O
0,3--------->0,15
=> \(V_{dd.Ba\left(OH\right)_2}=\dfrac{0,15}{0,5}=0,3\left(l\right)=300\left(ml\right)\)
a, \(n_{KOH}=\dfrac{8,4}{56}=0,15\left(mol\right)\)
PT: \(CH_3COOH+KOH\rightarrow CH_3COOK+H_2O\)
Theo PT: \(n_{CH_3COOH}=n_{KOH}=0,15\left(mol\right)\Rightarrow C_{M_{CH_3COOH}}=\dfrac{0,15}{0,5}=0,3\left(M\right)\)
b, \(n_{Na_2CO_3}=0,2.0,5=0,1\left(mol\right)\)
PT: \(2CH_3COOH+Na_2CO_3\rightarrow2CH_3COONa+CO_2+H_2O\)
Xét tỉ lệ: \(\dfrac{0,15}{2}< \dfrac{0,1}{1}\), ta được Na2CO3 dư.
Theo PT: \(n_{CO_2}=\dfrac{1}{2}n_{CH_3COOH}=0,075\left(mol\right)\Rightarrow V_{CO_2}=0,075.22,4=1,68\left(l\right)\)
\(Na_2CO_3(0,05)+2CH_3COOH(0,1)--->2CH_3COONa(0,1)+CO_2(0,05)+H_2O\)
\(m_{Na_2CO_3}=\dfrac{10,6.50}{100}=5,3\left(g\right)\)
\(\Rightarrow n_{Na_2CO_3}=0,05\left(mol\right)\)
Phản ứng vừa đủ:
Theo PTHH: \(nCH_3COOH=0,1\left(mol\right)\)
\(\Rightarrow m_{CH_3COOH}=60.0,1=6\left(g\right)\)
\(b)\)
Theo PTHH: \(n_{CO_2}=0,05\left(mol\right)\)
\(\Rightarrow m_{CO_2}=2,2\left(g\right)\)
\(m_{ddsau}=50+50-2,2=97,8\left(g\right)\)
Dung dich muối tạo thành là CH3COONa
Theo PTHH: \(n_{CH_3COONa}=0,1\left(mol\right)\)
\(\Rightarrow m_{CH_3COONa}=82.0,1=8,2\left(g\right)\)
\(\Rightarrow C\%CH_3COONa=\dfrac{8,2}{97,8}.100\%=8,38\%\)
\(n_{CH_3COOH}=0.5\cdot1=0.5\left(mol\right)\)
\(CH_3COOH+NaOH\rightarrow CH_3COONa+H_2O\)
\(0.5...........................................0.5\)
\(m_{CH_3COOH}=0.5\cdot60=30\left(g\right)\)
\(m_{CH_3COONa}=0.5\cdot82=41\left(g\right)\)
\(n_{Na_2CO_3}=0.2\cdot0.5=0.1\left(mol\right)\)
\(2CH_3COOH+Na_2CO_3\rightarrow2CH_3COONa+CO_2+H_2O\)
\(0.2.........................0.1\)
\(\Rightarrow CH_3COOHdư\)
\(n_{CO_2}=n_{Na_2CO_3}=0.1\left(mol\right)\)
\(V_{CO_2}=0.1\cdot22.4=2.24\left(l\right)\)