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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
=> \(V_{dd.NaOH}=\dfrac{0,02}{0,2}=0,1\left(l\right)=100\left(ml\right)\)
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
200ml = 0,2l
\(n_{Zn}=\dfrac{19,5}{65}=0,3\left(mol\right)\)
Pt : \(Zn+2HCl\rightarrow ZnCl_2+H_2|\)
1 2 1 1
0,3 0,6 0,3 0,3
a) \(n_{ZnCl2}=\dfrac{0,3.1}{1}=0,3\left(mol\right)\)
\(C_{M_{ZnCl2}}=\dfrac{0,3}{0,2}=1,5\left(M\right)\)
b) \(n_{H2}=\dfrac{0,3.1}{1}=0,3\left(mol\right)\)
\(V_{H2\left(dktc\right)}=0,3.22,4=6,72\left(l\right)\)
c) Pt : \(NaOH+HCl\rightarrow NaCl+H_2O|\)
1 1 1 1
0,6 0,6
\(n_{NaOH}=\dfrac{0,6.1}{1}=0,6\left(mol\right)\)
\(m_{NaOH}=0,6.40=24\left(g\right)\)
\(m_{ddNaOH}=\dfrac{24.100}{20}=120\left(g\right)\)
Chúc bạn học tốt
\(n_{Mg}=\dfrac{6}{24}=0,25\left(mol\right)\)
a) Pt : \(Mg+2HCl\rightarrow MgCl_2+H_2|\)
1 2 1 1
0,25 0,5 0,25
b) \(n_{H2}=\dfrac{0,25.1}{1}=0,25\left(mol\right)\)
\(V_{H2\left(dktc\right)}=0,25.22,4=5,6\left(l\right)\)
c) \(n_{HCl}=\dfrac{0,25.2}{1}=0,5\left(mol\right)\)
\(V_{ddHCl}=\dfrac{0,5}{2}=0,25\left(l\right)=250\left(ml\right)\)
Chúc bạn học tốt
\(a,PTHH:Mg+2HCl\rightarrow MgCl_2+H_2\\ b,n_{Mg}=\dfrac{6}{24}=0,25\left(mol\right)\\ \Rightarrow n_{H_2}=0,25\left(mol\right)\\ \Rightarrow V_{H_2\left(đktc\right)}=0,25\cdot22,4=5,6\left(l\right)\\ c,n_{HCl}=2n_{Mg}=0,5\left(mol\right)\\ \Rightarrow V_{dd_{HCl}}=\dfrac{0,5}{2}=0,25\left(l\right)\)
nKOH = 0,5.0,3 = 0,15 mol
CH3COOH + KOH → CH3COOK + H2O
0,15 0,15 0,15 mol
a) CM CH3COOH = 0,15/0,2 =0,75M
b) Thể tích của dung dịch thu được sau phản ứng: 500 ml
CM CH3COOK = 0,15/0,5 = 0,3M
c) Phản ứng lên men giấm
C2H5OH + O2 → CH3COOH + H2O
0,15 0,15
→ mC2H5OH = 0,15.46 = 6,9 gam
\(n_{KOH}=0,5\cdot0,3=0,15mol\)
\(CH_3COOH+KOH\rightarrow CH_3COOK+H_2O\)
0,15 0,15 0,15 0,15
a)\(C_{M_{CH_3COOH}}=\dfrac{0,15}{0,2}=0,75M\)
b)\(C_{M_{CH_3COOK}}=\dfrac{0,15}{0,2+0,3}=0,3M\)
a, \(n_{Mg}=\dfrac{2,4}{24}=0,1\left(mol\right)\)
PT: \(Mg+2CH_3COOH\rightarrow\left(CH_3COO\right)_2Mg+H_2\)
Theo PT: \(n_{H_2}=n_{Mg}=0,1\left(mol\right)\Rightarrow V_{H_2}=0,1.24,79=2,479\left(l\right)\)
b, \(n_{\left(CH_3OO\right)_2Mg}=n_{Mg}=0,1\left(mol\right)\)
Ta có: m dd sau pư = 2,4 + 100 - 0,1.2 = 102,2 (g)
\(\Rightarrow C\%_{\left(CH_3COO\right)_2Mg}=\dfrac{0,1.142}{102,2}.100\%\approx13,89\%\)
c, Bạn bổ sung thêm CM của NaOH nhé.
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
=> \(V_{ddNaOH}=\dfrac{0,2}{0,5}=0,4\left(l\right)\)
\(a,Zn+2CH_3COOH\rightarrow\left(CH_3COO\right)_2Zn+H_2\\ n_{H_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\\ b,n_{CH_3COOH}=2.0,2=0,4\left(mol\right)\\ C_{MddCH_3COOH}=\dfrac{0,4}{0,4}=1\left(M\right)\\ c,CH_3COOH+KOH\rightarrow CH_3COOK+H_2O\\ n_{CH_3COOK}=n_{CH_3COOH}=0,4\left(mol\right)\\ V_{ddCH_3COOK}=400+400=800\left(ml\right)=0,8\left(l\right)\\ C_{MddCH_3COOK}=\dfrac{0,4}{0,8}=0,5\left(M\right)\)
\(Zn+2CH_3COOH\rightarrow\left(CH_3COO\right)_2Zn+H_2\uparrow\)
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+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
=> \(V_{dd.NaOH}=\dfrac{0,02}{0,75}=\dfrac{2}{75}\left(l\right)=\dfrac{80}{3}\left(ml\right)\)