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\(a,n_{CH_3COOH}=\dfrac{120.20}{100}=24\left(g\right)\\ \rightarrow n_{CH_3COOH}=\dfrac{24}{60}=0,4\left(mol\right)\)
PTHH: \(2CH_3COOH+Na_2CO_3\rightarrow2CH_3COONa+CO_2\uparrow+H_2O\)
0,4----------->0,2-------------->0,4-------------->0,2
\(\rightarrow m_{ddNa_2CO_3}=\dfrac{0,2.106}{10\%}=212\left(g\right)\)
\(\rightarrow m_{ddA}=212+120-0,2.44=323,2\left(g\right)\\ \rightarrow C\%_{CH_3COONa}=\dfrac{0,4.82}{323,2}.100\%=10,15\%\)
b, PTHH: \(C_2H_5OH+O_2\underrightarrow{\text{men giấm}}CH_3COOH+H_2O\)
0,4<------------------------0,4
\(\rightarrow V_{ddC_2H_5OH}=\dfrac{0,4.46.100}{0,8.46}=50\left(ml\right)\)
a,VddC2H5OH=23+27=50(ml)
Đr=\(\dfrac{23}{50}\).100=46o
b,mC2H5OH=23.0,8=18,4(g)
nC2H5OH=\(\dfrac{18,4}{46}\)=0,4(mol)
PTHH: 2C2H5OH + 2K ---> 2C2H5OK + H2
0,4-------------------------------------->0,2
=> VH2 = 0,2.24 = 4,8 (l
a,VddC2H5OH=23+27=50(ml)
Đr=2350.100=46ob, mC2H5OH=23.0,8=18,4(g)
nC2H5OH=18,446=0,4(mol)a,
VddC2H5OH=23+27=50(ml)Đr=2350.100=46ob,
mC2H5OH=23.0,8=18,4(g)
nC2H5OH=18,446=0,4(mol)
PTHH: 2C2H5OH + 2K ---> 2C2H5OK + H2
0,4-------------->0,2
=> VH2 = 0,2.24 = 4,8 (l)
\(a,V_{ddC_2H_5OH}=23+27=50\left(ml\right)\\ Đ_r=\dfrac{23}{50}.100=46^o\\ b,m_{C_2H_5OH}=23.0,8=18,4\left(g\right)\\ n_{C_2H_5OH}=\dfrac{18,4}{46}=0,4\left(mol\right)\)
PTHH: 2C2H5OH + 2K ---> 2C2H5OK + H2
0,4-------------------------------------->0,2
=> VH2 = 0,2.24 = 4,8 (l)
a, \(C_2H_6O+3O_2\underrightarrow{t^o}2CO_2+3H_2O\)
b, \(n_{CO_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\)
Theo PT: \(n_{O_2}=\dfrac{3}{2}n_{CO_2}=0,3\left(mol\right)\)
\(\Rightarrow V_{O_2}=0,3.22,4=6,72\left(l\right)\)
c, \(n_{C_2H_6O}=\dfrac{1}{2}n_{CO_2}=0,1\left(mol\right)\)
\(\Rightarrow m_{C_2H_6O}=0,1.46=4,6\left(g\right)\)
\(\Rightarrow V_{C_2H_6O}=\dfrac{4,6}{0,8}=5,75\left(ml\right)\)
Độ rượu = \(\dfrac{5,75}{50}.100=11,5^o\)
\(a,PTHH:Zn+2HCl\rightarrow ZnCl_2+H_2\\ b,n_{Zn}=\dfrac{13}{65}=0,2\left(mol\right)\\ n_{HCl}=2\cdot0,15=0,3\left(mol\right)\)
Vì \(\dfrac{n_{Zn}}{1}>\dfrac{n_{HCl}}{2}\) nên sau p/ứ Zn dư
\(\Rightarrow n_{Zn}=\dfrac{1}{2}n_{HCl}=0,15\left(mol\right)\\ \Rightarrow m_{Zn}=0,15\cdot65=9,75\\ \Rightarrow m_{Zn\left(dư\right)}=13-9,75=3,25\left(g\right)\\ c,n_{H_2}=n_{Zn}=0,15\left(mol\right)\\ \Rightarrow V_{H_2}=0,15\cdot22,4=3,36\left(l\right)\)
\(a,V_{C_2H_5OH}=\dfrac{96.30}{100}=28,8\left(ml\right)\\ \rightarrow m_{C_2H_5OH}=28,8.0,8=23,04\left(ml\right)\\ \rightarrow n_{C_2H_5OH}=\dfrac{23,04}{46}=0,5\left(mol\right)\)
PTHH: 2C2H5OH + 2Na ---> 2C2H5ONa + H2
0,5----------------------------------->0,25
\(\rightarrow V_{H_2}=0,25.22,4=5,6\left(l\right)\)
b, \(n_{CH_3COOH}=\dfrac{36}{60}=0,6\left(mol\right)\)
PTHH: \(C_2H_5OH+CH_3COOH\xrightarrow[t^o]{H_2SO_{4\left(đ\right)}}CH_3COOC_2H_5+H_2O\)
LTL: 0,5 < 0,6 => CH3COOH dư
Theo pthh: nCH3COOH = nC2H5OH = 0,5 (mol)
=> meste = 0,5.88.70% = 30,8 (g)