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a.b.\(n_{CO_2}=\dfrac{5,6}{22,4}=0,25mol\)
\(2CH_3COOH+Na_2CO_3\rightarrow2CH_3COONa+CO_2+H_2O\)
0,5 0,25 ( mol )
\(m_{CH_3COOH}=0,5.60=30g\)
\(\%m_{CH_3COOH}=\dfrac{30}{45}.100=66,67\%\)
\(\%m_{C_2H_5OH}=100\%-66,67\%=33,33\%\)
c.\(m_{NaOH}=50.20\%=10g\)
\(n_{NaOH}=\dfrac{10}{40}=0,25mol\)
\(2NaOH+CO_2\rightarrow Na_2CO_3+H_2O\)
\(\dfrac{0,25}{2}\) < \(\dfrac{0,25}{1}\) ( mol )
0,25 0,125 ( mol )
\(m_{Na_2CO_3}=0,125.106=13,25g\)
a)
2CH3COOH + Na2CO3 --> 2CH3COONa + CO2 + H2O
b) \(n_{CO_2}=\dfrac{5,6}{22,4}=0,25\left(mol\right)\)
PTHH: 2CH3COOH + Na2CO3 --> 2CH3COONa + CO2 + H2O
0,5<-----------------------------------0,25
=> mCH3COOH = 0,5.60 = 30 (g)
=> mC2H5OH = 45 - 30 = 15 (g)
c) \(n_{NaOH}=\dfrac{50.20\%}{40}=0,25\left(mol\right)\)
Xét tỉ lệ: \(\dfrac{n_{NaOH}}{n_{CO_2}}=\dfrac{0,25}{0,25}=1\) => Tạo muối NaHCO3
PTHH: NaOH + CO2 --> NaHCO3
0,25-------------->0,25
=> mNaHCO3 = 0,25.84 = 21 (g)
Câu 1:
Gọi \(\left\{{}\begin{matrix}n_{CH3COOH}:a\left(mol\right)\\n_{C2H5OH}:b\left(mol\right)\end{matrix}\right.\)
PTHH:
\(2Na+2CH_3COOH\rightarrow2CH_3COONa+H_2\)
\(2Na+2C_2H_5OH\rightarrow2C_2H_5ONa+H_2\)
Giải hệ PT:
\(\left\{{}\begin{matrix}60a+46b=27,2\\a+b=0,5\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}a=0,3\\b=0,2\end{matrix}\right.\)
\(\Rightarrow n_{CH3COOC2H5}=n_{C2H5OH}=0,2\left(mol\right)\)
\(\Rightarrow m_{C2H5COOC2H5}=0,2.88=17,6\left(g\right)\)
Câu 2:
a, PTHH:
\(2Na+2CH_3COOH\rightarrow2CH_3COONa+H_2\)
\(2Na+2C_2H_5OH\rightarrow2C_2H_5ONa+H_2\)
\(CaCO_3+2CH_3COOH\rightarrow\left(CH_3COO\right)_2Ca+H_2O+CO_2\)
b, Ta có:
\(n_{CO2}=\frac{2,24}{22,4}=0,1\left(mol\right)\)
\(n_{H2}=\frac{5,6}{22,4}=0,25\left(mol\right)\)
\(\Rightarrow n_{CH3COOH}=2n_{CO2}=0,1.2=0,2\left(mol\right)\)
\(\Rightarrow n_{Hh}=0,25.2=0,5\left(mol\right)\Rightarrow n_{C2H5OH}=0,5-0,2=0,3\left(mol\right)\)
\(\Rightarrow m=25,8\left(g\right)\)
c,\(PTHH:C_2H_5OH+CH_3COOH\rightarrow CH_3COOC_2H_5+H_2O\)
\(\Rightarrow n_{CH3COOC2H5}=n_{CH3COOH}=0,2\left(mol\right)\)
\(\Rightarrow m_{CH3COOC2H5}=14,08\left(g\right)\)
a)
C2H5OH + Na → C2H5ONa + 1/2 H2
CH3COOH + Na → CH3COONa + 1/2 H2
b)
Theo PTHH :
n C2H5OH = a(mol) ; n CH3COOH = b(mol)
=> 46a + 60b = 12,9(1)
n H2 = 0,5a + 0,5b = 2,8/22,5 = 0,125(2)
Từ (1)(2) suy ra a = 0,15 ; b = 0,1
%m C2H5OH = 0,15.46/12,9 .100% = 53,49%
%m CH3COOH = 100% -53,49% = 46,51%
b)
n C2H5ONa = a = 0,15 mol
n CH3COONa = b = 0,1(mol)
=> m muối = 0,15.69 + 0,1.82 = 18,55 gam
nH2 = 0,2 mol
Gọi x, y lần lượt là số mol của C2H5OH, CH3COOH ( x,y > 0 )
2C2H5OH + 2Na \(\rightarrow\) 2C2H5ONa + H2
x...................x................x................0,5x
2CH3COOH + 2Na \(\rightarrow\) 2CH3COONa + H2
y.........................y.................y................0,5y
Ta có hệ
\(\left\{{}\begin{matrix}46x+60y=21,2\\0,5x+0,5y=0,2\end{matrix}\right.\)
\(\Rightarrow\) \(\left\{{}\begin{matrix}x=0,2\\y=0,2\end{matrix}\right.\)
\(\Rightarrow\) %C2H5OH = \(\dfrac{0,2.46.100\%}{21,2}\)\(\approx\) 43,4%
\(\Rightarrow\) %CH3COOH = \(\dfrac{0,2.60.100\%}{21,2}\) \(\approx\) 56,6%
a. C2H5OH + Zn ---/----> Ko xảy ra
2CH3COOH + Zn --------> (CH3COOH)2Zn + H2
b. Ta có :
n hỗn hợp = 6,72/22,4=0,3 mol
Theo PTHH: n CH3COOH = 2 n H2 =0,6 mol
=> m CH3COOH = 0,6.60=36g
=> m C2H5OH =60-36=24g
=> % m C2H5OH= 24/60.100%=40%
=>% m CH3COOH = 100%-40%=60%
c. Ta có:
V rượu = 24/0,8=30 ml
=>V dd= 30.100%/15*=200ml
\(a)\)11,6 gam hỗn hợp: \(\left\{{}\begin{matrix}CH_3COOH:a\left(mol\right)\\C_2H_5OH:b\left(mol\right)\end{matrix}\right.\)
\(\Rightarrow60a+46b=11,6\)\((I)\)
\(2CH_3 COOH(a)+2Na--->2CH_3COONa+H_2(0,5a)\)
\(2C_2H_5OH\left(b\right)+2Na--->2C_2H_5ONa+H_2\left(0,5b\right)\)
\(n_{H_2}=0,12\left(mol\right)\)
\(\Rightarrow0,5a+0,5b=0,12\)\(\left(II\right)\)
Từ (I) và (II) \(\Rightarrow\left\{{}\begin{matrix}a=0,04\\b=0,2\end{matrix}\right.\)
\(\Rightarrow\%m_{CH_3COOH}=\dfrac{0,04.60}{11,6}.100\%=20,69\%\)
\(\Rightarrow\%m_{C_2H_5OH}=79,31\%\)
\(b)\)
Sau phản ứng thu được: \(\left\{{}\begin{matrix}CH_3COONa:a=0,04\left(mol\right)\\C_2H_5ONa:b=0,2\left(mol\right)\\H_2:0,5a+0,5b=0,12\left(mol\right)\\Na\left(dư\right)\end{matrix}\right.\)
Khối lượng Na không tính được
\(\Rightarrow m_{CH_3COONa}=3,28\left(g\right)\)
\(m_{C_2H_5ONa}=13,6\left(g\right)\)
\(m_{H_2}=0,24\left(g\right)\)
P/s: Đề ra chưa được hoàn hảo, nên dùng lượng vừa đủ Na. =]]
a) Gọi số mol CH3COOH, C2H5OH là a, b (mol)
=> 60a + 46b = 25,8 (1)
\(n_{H_2}=\dfrac{5,6}{22,4}=0,25\left(mol\right)\)
PTHH: 2Na + 2CH3COOH --> 2CH3COONa + H2
a------------------------->0,5a
2Na + 2C2H5OH --> 2C2H5ONa + H2
b--------------------->0,5b
=> 0,5a + 0,5b = 0,25 (2)
(1)(2) => a = 0,2 (mol); b = 0,3 (mol)
=> \(\left\{{}\begin{matrix}\%m_{CH_3COOH}=\dfrac{0,2.60}{25,8}.100\%=46,51\%\\\%m_{C_2H_5OH}=\dfrac{0,3.46}{25,8}.100\%=53,49\%\end{matrix}\right.\)
b)
\(n_{CH_3COOC_2H_5}=\dfrac{13,2}{88}=0,15\left(mol\right)\)
PTHH: CH3COOH + C2H5OH --H2SO4(đ),to--> CH3COOC2H5 + H2O
Xét tỉ lệ: \(\dfrac{0,2}{1}< \dfrac{0,3}{1}\) => Hiệu suất tính theo CH3COOH
PTHH: CH3COOH + C2H5OH --H2SO4(đ),to--> CH3COOC2H5 + H2O
0,15<---------------------------------0,15
=> \(H=\dfrac{0,15}{0,2}.100\%=75\%\)
CH3COOH +Na =>CH3COONa +1/2H2
x mol =>0,5x mol
C2H5OH+ Na=>C2H5ONa +1/2H2
y mol. =>0,5y mol
nH2=5,6/22,4=0,25 mol=>x+y=0,5
mhh bđ=60x+46y=27,2
=>x=0,3 và y=0,2
mCH3COOH=0,3.60=18g=>%mCH3COOH=18/27,2.100%=66,17%
%mC2H5OH=33,82%