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\(n_{CH_3COOH}=0.5\cdot1=0.5\left(mol\right)\)
\(CH_3COOH+NaOH\rightarrow CH_3COONa+H_2O\)
\(0.5...........................................0.5\)
\(m_{CH_3COOH}=0.5\cdot60=30\left(g\right)\)
\(m_{CH_3COONa}=0.5\cdot82=41\left(g\right)\)
\(n_{Na_2CO_3}=0.2\cdot0.5=0.1\left(mol\right)\)
\(2CH_3COOH+Na_2CO_3\rightarrow2CH_3COONa+CO_2+H_2O\)
\(0.2.........................0.1\)
\(\Rightarrow CH_3COOHdư\)
\(n_{CO_2}=n_{Na_2CO_3}=0.1\left(mol\right)\)
\(V_{CO_2}=0.1\cdot22.4=2.24\left(l\right)\)
a) nCH3COOH= 0,4(mol)
PTHH: CH3COOH + NaOH -> CH3COONa + H2O
0,4____________0,4(mol)
=> mNaOH=0,4. 40=16(g)
b) nCH3COOH= 1(mol)
nC2H5OH= 100/46= 50/23(mol)
Vì : 1/1< 50/23 :1
=> C2H5OH dư, CH3COOH hết, tính theo nCH3COOH.
PTHH: CH3COOH + C2H5OH \(⇌\) CH3COOC2H5 + H2O (đk: H+ , nhiệt độ)
Ta có: nCH3COOC2H5(thực tế)= 0,625(mol)
Mà theo LT: nCH3COOC2H5(LT)= nCH3COOH=1(mol)
=>H= (0,625/1).100=62,5%
PTHH: \(Na_2CO_3+2CH_3COOH\rightarrow2CH_3COONa+H_2O+CO_2\uparrow\)
Tính theo sản phẩm
Ta có: \(\left\{{}\begin{matrix}\Sigma n_{CH_3COOH}=\dfrac{160\cdot15\%}{60}=0,4\left(mol\right)\\n_{CO_2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}n_{CH_3COOH\left(dư\right)}=0,2\left(mol\right)=n_{CH_3COONa}\\n_{Na_2CO_3}=0,1\left(mol\right)\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}m_{Na_2CO_3}=0,1\cdot106=10,6\left(g\right)\\m_{CH_3COONa}=0,2\cdot82=16,4\left(g\right)\\m_{CH_3COOH\left(dư\right)}=0,2\cdot60=12\left(g\right)\\m_{CO_2}=0,1\cdot44=4,4\left(g\right)\end{matrix}\right.\)
Mặt khác: \(m_{dd\left(sau.p/ứ\right)}=m_{Na_2CO_3}+m_{ddAxit}-m_{CO_2}=166,2\left(g\right)\)
\(\Rightarrow\left\{{}\begin{matrix}C\%_{CH_3COONa}=\dfrac{16,4}{166,2}\cdot100\%\approx9,87\%\\C\%_{CH_3COOH\left(dư\right)}=\dfrac{12}{166,2}\cdot100\%\approx7,22\%\end{matrix}\right.\)
Ta có: \(\left\{{}\begin{matrix}n_{CO_2}=0,1\left(mol\right)\\n_{NaOH}=0,5\cdot0,3=0,15\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\) Tạo 2 muối
PTHH: \(CO_2+NaOH\rightarrow NaHCO_3\)
a_______a__________a (mol)
\(CO_2+2NaOH\rightarrow Na_2CO_3+H_2O\)
b_______2b_________2b (mol)
Ta lập được HPT \(\left\{{}\begin{matrix}a+b=0,1\\a+2b=0,15\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}a=0,05\\b=0,05\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}m_{NaHCO_3}=0,05\cdot84=4,2\left(g\right)\\m_{Na_2CO_3}=0,05\cdot106=5,3\left(g\right)\end{matrix}\right.\)
\(m_{CH_3COOH}=24\%.150=36\left(g\right)\\ \rightarrow n_{CH_3COOH}=\dfrac{36}{60}=0,6\left(mol\right)\)
PTHH: 2CH3COOH + Na2CO3 ---> 2CH3COONa + CO2 + H2O
0,6 0,3 0,6 0,3
=> VCO2 = 0,3.22,4 = 6,72 (l)
\(m_{Na_2CO_3}=0,3.31,8\left(g\right)\)
=> \(m_{ddNa_2CO_3}=\dfrac{31,8}{21,2\%}=150\left(g\right)\)
mCO2 = 0,3.44 = 13,2 (g)
\(m_{dd}=150+150-13,2=286,8\left(g\right)\)
\(m_{CH_3COONa}=0,3.82=24,6\left(g\right)\\ \rightarrow C\%_{CH_3COONa}=\dfrac{24,6}{286,8}=8,58\%\)
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: \(n_{CO_2}=\dfrac{1}{2}n_{CH_3COOH}=0,075\left(mol\right)\Rightarrow V_{CO_2}=0,075.22,4=1,68\left(l\right)\)
Câu 8 :
\(n_{MgCO3}=\dfrac{42}{84}=0,5\left(mol\right)\)
Pt : \(2CH_3COOH+MgCO_3\rightarrow\left(CH_3COO\right)_2Mg+CO_2+H_2O\)
1 0,5 0,5
a) \(V_{CO2\left(dktc\right)}=0,5.22,4=11,2\left(l\right)\)
b) \(V_{CH3COOH}=\dfrac{1}{2}=0,5\left(l\right)\)
c) Pt : \(CH_3COOH+NaOH\rightarrow CH_3COONa+H_2O\)
1 1
300ml = 0,3l
\(C_{MCH3COONa}=\dfrac{1}{0,3}=\dfrac{10}{3}\left(M\right)\)
Chúc bạn học tốt
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
=> \(V_{dd.Ba\left(OH\right)_2}=\dfrac{0,15}{0,5}=0,3\left(l\right)=300\left(ml\right)\)
\(n_{CH_3COOH}=0.5\cdot1=0.5\left(mol\right)\)
\(CH_3COOH+NaOH\rightarrow CH_3COONa+H_2O\)
\(0.5.....................0.5..................0.5\)
\(m_{CH_3COOH}=0.5\cdot60=30\left(g\right)\)
\(m_{CH_3COONa}=0.5\cdot82=41\left(g\right)\)
\(m_{NaOH}=0.5\cdot40=20\left(g\right)\)
\(n_{Na_2CO_3}=0.1\cdot0.5=0.05\left(mol\right)\)
\(2CH_3COOH+Na_2CO_3\rightarrow2CH_3COONa+CO_2+H_2O\)
\(2.........................1\)
\(0.5.......................0.05\)
\(LTL:\dfrac{0.5}{2}>\dfrac{0.05}{1}\Rightarrow CH_3COOHdư\)
\(V_{CO_2}=0.05\cdot22.4=1.12\left(l\right)\)