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a) \(n_{\left(CH_3COO\right)_2Mg}=\dfrac{1,42}{142}=0,01\left(mol\right)\)
PTHH: Mg + 2CH3COOH --> (CH3COO)2Mg + H2
0,02<-----------0,01-------->0,01
=> VH2 = 0,01.22,4 = 0,224 (l)
\(C_{M\left(CH_3COOH\right)}=\dfrac{0,02}{0,2}=0,1M\)
b)
PTHH: CH3COOH + NaOH --> CH3COONa + H2O
0,02------>0,02
=> \(V_{dd.NaOH}=\dfrac{0,02}{0,2}=0,1\left(l\right)=100\left(ml\right)\)
a.\(n_{\left(CH_3COO\right)_2Mg}=\dfrac{14,2}{142}=0,1mol\)
\(2CH_3COOH+Mg\rightarrow\left(CH_3COO\right)_2Mg+H_2\)
0,2 0,1 0,1 ( mol )
\(C_{M_{CH_3COOH}}=\dfrac{0,2}{0,25}=0,8M\)
\(V_{H_2}=0,1.22,4=2,24l\)
b.\(NaOH+CH_3COOH\rightarrow CH_3COONa+H_2O\)
0,2 0,2 ( mol )
\(V_{NaOH}=\dfrac{0,2}{0,5}=0,4l\)
\(n_{\left(CH_3COO\right)_2Mg}=\dfrac{14,2}{142}=0,1\left(mol\right)\)
PTHH: 2CH3COOH + Mg ---> (CH3COO)2Mg + H2
0,2<---------------------------0,1---------->0,1
=> \(\left\{{}\begin{matrix}C_{M\left(CH_3COOH\right)}=\dfrac{0,2}{0,25}=0,8M\\V_{H_2}=0,1.22,4=2,4\left(l\right)\end{matrix}\right.\)
PTHH: CH3COOH + NaOH ---> CH3COONa + H2O
0,2------------->0,2
=> \(V_{ddNaOH}=\dfrac{0,2}{0,5}=0,4\left(l\right)\)
\(a,n_{\left(CH_3COO\right)_2Mg}=\dfrac{7,1}{142}=0,05\left(mol\right)\)
PTHH: \(Mg+2CH_3COOH\rightarrow\left(CH_3COO\right)_2Mg+H_2\uparrow\)
0,1<----------------0,05-------------->0,05
\(\rightarrow V_{H_2}=0,05.22,4=1,12\left(l\right)\\ b,C\%_{CH_3COOH}=\dfrac{0,1.60}{100}.100\%=6\%\)
\(c,n_{C_2H_5OH}=\dfrac{6,9}{46}=0,15\left(mol\right)\)
PTHH: \(CH_3COOH+C_2H_5OH\xrightarrow[H_2SO_{4\left(đặc\right)}]{t^o}CH_3COOC_2H_5+H_2O\)
bđ 0,1 0,15
pư 0,1 0,1
spư 0 0,05 0,1
\(\rightarrow m_{este}=0,1.80\%.88=7,04\left(g\right)\)
\(n_{\left(CH_3COO\right)_2Mg}=\dfrac{2,84}{142}=0,02\left(mol\right)\)
PTHH :
\(Mg+2CH_3COOH\rightarrow\left(CH_3COO\right)_2Mg+H_2\uparrow\)
0,04 0,02 0,02
\(a,C_M=\dfrac{n}{V}=\dfrac{0,04}{0,1}=0,4M\)
\(b,V_{H_2}=0,02.22,4=0,448\left(l\right)\)
\(c,PTHH:\)
\(CH_3COOH+C_2H_5OH\underrightarrow{t^o,H_2SO_{4\left(đ\right)}}CH_3COOC_2H_5+H_2O\)
0,04 0,04
\(m_{este}=0,04.90\%.88=3,168\left(g\right)\)
a) \(n_{\left(CH_3COO\right)_2Mg}=\dfrac{1,42}{142}=0,01\left(mol\right)\)
PTHH: Mg + 2CH3COOH --> (CH3COO)2Mg + H2
0,02<------------0,01----->0,01
=> \(C_{M\left(dd.CH_3COOH\right)}=\dfrac{0,02}{0,05}=0,4M\)
b) VH2 = 0,01.22,4 = 0,224 (l)
đổi 500ml = 0,5l
n(CH\(_3\)COOH)\(_2\)Mg= \(\dfrac{14,2}{142}\)= 0,1mol
2CH3COOH + Mg \(\rightarrow\) (CH3COO)2Mg + H2
0,2mol 0,1mol 0,1mol
a/ CCH3COOH= \(\dfrac{0,2}{0,5}\)=0,4M
b/ VH\(_2\) 0,1 . 22,4 = 2,24l
c/ nCH\(_3\)COOH= 0,2mol
CH3COOH + NaOH \(\rightarrow\) CH3COONa + H2
0,2mol 0,2mol
V\(_{dd_{NaOH}}\)= \(\dfrac{0,2}{0,5}\)= 0,4l
2CH3COOH+Mg->(CH3COO)2Mg+H2
0,02---------------0,01-------0,01----------0,01
n muối=0,01mol
=>CM=\(\dfrac{0,02}{0,04}=0,5M\)
=>VH2=0,01.22,4=0,224l
CH3COOH+NaOH->CH3COONa+H2O
0,02--------------0,02
=>VNaOH=\(\dfrac{0,02}{0,75}=0,03l\)
a) \(n_{\left(CH_3COO\right)_2Mg}=\dfrac{1,42}{142}=0,01\left(mol\right)\)
PTHH: Mg + 2CH3COOH --> (CH3COO)2Mg + H2
0,01<-------0,02<------------0,01------->0,01
=> \(C_{M\left(dd.CH_3COOH\right)}=\dfrac{0,02}{0,04}=0,5M\)
b) VH2 = 0,01.22,4 = 0,224 (l)
c)
PTHH: NaOH + CH3COOH --> CH3COONa + H2O
0,02<------0,02
=> \(V_{dd.NaOH}=\dfrac{0,02}{0,75}=\dfrac{2}{75}\left(l\right)=\dfrac{80}{3}\left(ml\right)\)
\(a,Zn+2CH_3COOH\rightarrow\left(CH_3COO\right)_2Zn+H_2\\ n_{H_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\\ b,n_{CH_3COOH}=2.0,2=0,4\left(mol\right)\\ C_{MddCH_3COOH}=\dfrac{0,4}{0,4}=1\left(M\right)\\ c,CH_3COOH+KOH\rightarrow CH_3COOK+H_2O\\ n_{CH_3COOK}=n_{CH_3COOH}=0,4\left(mol\right)\\ V_{ddCH_3COOK}=400+400=800\left(ml\right)=0,8\left(l\right)\\ C_{MddCH_3COOK}=\dfrac{0,4}{0,8}=0,5\left(M\right)\)
\(Zn+2CH_3COOH\rightarrow\left(CH_3COO\right)_2Zn+H_2\uparrow\)
0,4 0,2
\(n_{H_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\)
\(b,C_{M_{CH_3COOH}}=\dfrac{0,4}{0,4}=1M\)
\(c,CH_3COOH+KOH\rightarrow CH_3COOK+H_2O\)
0,4 0,4 0,4
\(C_{M_{CH_3COOK}}=\dfrac{0,4}{0,4}=1M\)
PTHH: 2CH3COOH+Mg → (CH3COO)2Mg+H2
n(CH3COO)2Mg=2,13/142=0,015mol
=> nCH3COOH=0,015.2=0,03mol
=> CM(CH3COOH)=0,03/0,075=0,4M
=> nH2=n(CH3COO)2Mg=0,015mol
⇒VH2=0,015.22,4=0,336l
PTHH : CH3COOH+NaOH → CH3COONa+H2O
=> nNaOH=nCH3COOH=0,03mol
=> VNaOH=0,03/0,5=0,06l=60ml
2CH3COOH+Zn->(CH3COO)2Zn+H2
0,074-------------------------0,037----0,037
n (CH3COO)2Zn=0,037 mol
=>CM =\(\dfrac{0,074}{0,025}\)=3M
=>VH2=0,037.22,4=0,8288l