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\(n_{\left(CH_3COO\right)_2Mg}=\dfrac{0,71}{142}=0,005\left(mol\right)\)
PTHH: 2CH3COOH + Mg ---> (CH3COO)2Mg + H2
0,01<----------------------0,005---------->0,005
\(C_{M\left(CH_3COOH\right)}=\dfrac{0,01}{0,025}=4M\)
PTHH: CH3COOH + NaOH ---> CH3COONa + H2O
0,005--------->0,005
\(V_{ddNaOH}=\dfrac{0,005}{0,75}=\dfrac{1}{150}M\\ V_{H_2}=0,005.22,4=0,0112\left(l\right)\)
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)
CH3COOH + Mg ---> CH3COOMg + 1/2H2
(mol) 0,026 0,026 0,013
a) nCH3COOMg = 2,13 : 83 = 0,026 mol
=> C\(_M\)CH3COOH = 0,026 : 0,02 = 1,3 M
b) V\(_{H2}\)= 0,013 . 22,4 = 0,2912(lit)
c) CH3COOH + NaOH ----> CH3COONa + H2O
\(n_{\left(CH_3COO\right)_2Mg}=\dfrac{2,84}{142}=0,02\left(mol\right)\)
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) PTHH: \(Fe+2HCl\rightarrow FeCl_2+H_2\)
b) Ta có: \(\left\{{}\begin{matrix}n_{Fe}=\dfrac{11,2}{56}=0,2\left(mol\right)\\n_{HCl}=\dfrac{200\cdot10\%}{36,5}=\dfrac{40}{73}\left(mol\right)\end{matrix}\right.\)
Xét tỉ lệ: \(\dfrac{0,2}{1}< \dfrac{\dfrac{40}{73}}{2}\) \(\Rightarrow\) HCl còn dư, Fe phản ứng hết
\(\Rightarrow n_{H_2}=0,2mol\) \(\Rightarrow V_{H_2}=0,2\cdot22,4=4,48\left(l\right)\)
c) PTHH: \(HCl+NaOH\rightarrow NaCl+H_2O\)
Ta có: \(n_{HCl\left(dư\right)}=\dfrac{54}{365}\left(mol\right)=n_{NaOH}\)
\(\Rightarrow V_{NaOH}=\dfrac{\dfrac{54}{365}}{0,5}\approx0,3\left(l\right)=300\left(ml\right)\)
\(PTHH:2CH_3COOH+Ca\rightarrow\left(CH_3COO\right)_2Na+H_2\)
Ta có:
\(n_{\left(CH3COO\right)2Na}=\frac{4,47}{158}=0,03\left(mol\right)\)
\(\Rightarrow n_{CH3COOH}=0,03.2=0,06\left(mol\right)\)
\(\Rightarrow CM_{CH3COOH}=\frac{0,06}{0,2}=0,3M\)
\(n_{H2}=0,03\left(mol\right)\Rightarrow V_{H2}=0,03.22,4=6,72\left(l\right)\)
PTHH :
\(CH_3COOH+NaOH\rightarrow CH_3COONa+H_2O\)
_0,06________0,06_____________________
\(\Rightarrow V_{NaOH}=\frac{0,06}{0,5}=0,12\left(l\right)=120\left(ml\right)\)
a, \(2CH_3COOH+Ca\rightarrow\left(CH_3COO\right)_2Ca+H_2\)
\(n_{\left(CH_3COO\right)_2Ca}=\dfrac{3,95}{158}=0,025\left(mol\right)\)
Theo PT: \(n_{CH_3COOH}=2n_{\left(CH_3COO\right)_2Ca}=0,05\left(mol\right)\)
\(\Rightarrow C_{M_{CH_3COOH}}=\dfrac{0,05}{0,25}=0,2\left(M\right)\)
b, \(n_{H_2}=n_{\left(CH_3COO\right)_2Ca}=0,025\left(mol\right)\)
\(\Rightarrow V_{H_2}=0,025.22,4=0,56\left(l\right)\)
c, \(CH_3COOH+NaOH\rightarrow CH_3COONa+H_2O\)
\(n_{NaOH}=n_{CH_3COOH}=0,05\left(mol\right)\)
\(\Rightarrow V_{NaOH}=\dfrac{0,05}{0,5}=0,1\left(l\right)=100\left(ml\right)\)
Mong mọi người giúp mình ạ!!!