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a) PTHH: \(Mg+2CH_3COOH\rightarrow\left(CH_3COO\right)_2Mg+H_2\uparrow\)
Ta có: \(n_{CH_3COOH}=0,2\cdot1=0,2\left(mol\right)\)
\(\Rightarrow n_{Mg}=n_{H_2}=0,1\left(mol\right)\) \(\Rightarrow\left\{{}\begin{matrix}m_{Mg}=0,1\cdot24=2,4\left(g\right)\\V_{H_2}=0,1\cdot22,4=2,24\left(l\right)\end{matrix}\right.\)
b) PTHH: \(C_2H_5OH+O_2\xrightarrow[]{men}CH_3COOH+H_2O\)
Theo PTHH: \(n_{C_2H_5OH}=n_{CH_3COOH}=0,2\left(mol\right)\)
\(\Rightarrow m_{ddC_2H_5OH}=\dfrac{0,2\cdot46}{8\%}=115\left(g\right)\) \(\Rightarrow V_{C_2H_5OH}=\dfrac{115}{0,8}=143,75\left(ml\right)\)
Phương trình hóa học:
Mg + 2CH3COOH => (CH3COO)2Mg + H2
nCH3COOH = CM.V = 0.2 x 1 = 0.2 (mol)
Theo phương trình ==> nMg = 0.1 (mol) => mMg = n.M = 0.1 x 24 = 2.4 (g)
Theo phương trình ==> nH2 = 0.1 (mol) ==> VH2 =22.4 x n = 22.4 x 0.1 = 2.24 (l)
C2H5OH + O2 => (men giấm) CH3COOH + H2O
nCH3COOH = 0.2 (mol) => nC2H5OH = 0.2 (mol)
mC2H5OH = n.M = 0.2 x 46 = 9.2 (g)
V = m/D = 9.2/8 = 1.15ml
\(a)n_{CH_3COOH} = 0,2.2 = 0,4(mol)\\ Mg + 2CH_3COOH \to (CH_3COO)_2Mg + H_2\\ n_{Mg} = \dfrac{1}{2}n_{CH_3COOH} = 0,2(mol)\\ m_{Mg} = 0,2.24 = 4,8(gam)\\ b)\\ CH_3COOH + C_2H_5OH \buildrel{{H_2SO_4}}\over\rightleftharpoons CH_3COOC_2H_5 + H_2O\\ n_{CH_3COOH\ pư} = n_{este} = \dfrac{24,64}{88} = 0,28(mol)\\ H = \dfrac{0,28}{0,4}.100\% = 70\%\)
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)\(C_{M_{CH_3COOK}}=\dfrac{0,15}{0,2+0,3}=0,3M\)
Ta có: \(n_{HCl}=\dfrac{200}{1000}.2=0,4\left(mol\right)\)
\(PTHH:Mg+2HCl--->MgCl_2+H_2\uparrow\left(1\right)\)
a. Theo PT(1): \(n_{Mg}=n_{H_2}=n_{MgCl_2}=\dfrac{1}{2}.n_{HCl}=\dfrac{1}{2}.0,4=0,2\left(mol\right)\)
\(\Rightarrow\left\{{}\begin{matrix}m_{Mg}=0,2.24=4,8\left(g\right)\\V_{H_2}=0,2.22,4=4,48\left(lít\right)\end{matrix}\right.\)
b. \(PTHH:2NaOH+MgCl_2--->Mg\left(OH\right)_2\downarrow+2NaCl\left(2\right)\)
Ta có: \(n_{NaOH}=\dfrac{\dfrac{20\%.100}{100\%}}{40}=0,5\left(mol\right)\)
Ta thấy: \(\dfrac{0,5}{2}>\dfrac{0,2}{1}\)
Vậy NaOH dư.
Theo PT(2): \(n_{Mg\left(OH\right)_2}=n_{MgCl_2}=0,2\left(mol\right)\)
\(\Rightarrow m_{Mg\left(OH\right)_2}=0,2.58=11,6\left(g\right)\)
a: \(Mg+2HCl\rightarrow MgCl_2+H_2\)
200ml=0,2 lít
\(n_{HCl}=0.2\cdot22.4=4.48\left(mol\right)\)
\(\Leftrightarrow n_{H_2}=2.24\left(mol\right)\)
\(\Leftrightarrow m_{H_2}=n_{H_2}\cdot M=2.24\cdot1=2.24\left(g\right)\)
\(n_{MgCl_2}=2.24\left(mol\right)\)
\(\Leftrightarrow n_{Mg}=2.24\left(mol\right)\)
\(\Leftrightarrow m_{Mg}=2.24\cdot24=53.76\left(g\right)\)
a) nCH3COOH= 0,4(mol)
PTHH: CH3COOH + NaOH -> CH3COONa + H2O
0,4____________0,4(mol)
=> mNaOH=0,4. 40=16(g)
b) nCH3COOH= 1(mol)
nC2H5OH= 100/46= 50/23(mol)
Vì : 1/1< 50/23 :1
=> C2H5OH dư, CH3COOH hết, tính theo nCH3COOH.
PTHH: CH3COOH + C2H5OH \(⇌\) CH3COOC2H5 + H2O (đk: H+ , nhiệt độ)
Ta có: nCH3COOC2H5(thực tế)= 0,625(mol)
Mà theo LT: nCH3COOC2H5(LT)= nCH3COOH=1(mol)
=>H= (0,625/1).100=62,5%
a) n Zn = 6,5/65 = 0,1(mol)
Zn + 2CH3COOH $\to$ (CH3COO)2Zn + H2
Theo PTHH :
n CH3COOH = 2n Zn =0,2(mol)
C% CH3COOH = 0,2.60/200 .100% = 6%
b) n H2 = n Zn = 0,1(mol)
=> m dd sau pư = 6,5 + 200 - 0,1.2 = 206,3 gam
Theo PTHH : n (CH3COO)2Zn = n Zn = 0,1(mol)
=> C% (CH3COO)2Zn = 0,1.183/206,3 .100% = 8,87%
c)
C2H5OH + O2 $\xrightarrow{men\ giấm}$ CH3COOH + H2O
n C2H5OH pư = n CH3COOH = 0,2(mol)
=> m C2H5OH cần dùng = 0,2.46/80% = 11,5 gam
a) nZn=0,1(mol)
PTHH: Zn + 2 CH3COOH -> (CH3COO)2Zn + H2
0,1_______0,2_________0,1_____________0,1(mol)
mCH3COOH=0,2.60=12(g)
=> C%ddCH3COOH=(12/200).100=6%
b) mdd(CH3COO)2Zn= 6,5+200-0,1.2=206,3(g)
m(CH3COO)2Zn= 183 x 0,1=18,3(g)
=>C%dd(CH3COO)2Zn= (18,3/206,3).100=8,871%
c) C2H5OH + O2 -men giấm-> CH3COOH + H2O
nC2H5OH(LT)=nCH3COOH=0,2(mol)
=> nC2H5OH(TT)=0,2 : 80%= 0,25(mol)
=>mC2H5OH=0,25 x 46= 11,5(g)
a.b.\(n_{Mg}=\dfrac{4,8}{24}=0,2mol\)
\(2Mg+2CH_3COOH\rightarrow2\left(CH_3COO\right)_2Mg+H_2\)
0,2 0,2 0,2 ( mol )
\(C_{M_{CH_3COOH}}=\dfrac{0,2}{0,2}=1M\)
\(m_{\left(CH_3COO\right)_2Mg}=0,2.142=28,4g\)
c.Sửa đề: thu được 9,2g este
\(n_{CH_3COOC_2H_5}=\dfrac{9,2}{88}=0,1mol\)
\(CH_3COOH+C_2H_5OH\rightarrow CH_3COOC_2H_5+H_2O\)
Thực tế: 0,2 0,1 ( mol )
Lý thuyết: 0,1 0,1 ( mol )
\(H=\dfrac{0,1}{0,2}.100=50\%\)
a, \(n_{CH_3COOH}=0,2.1=0,2\left(mol\right)\)
PT: \(Mg+2CH_3COOH\rightarrow\left(CH_3COO\right)_2Mg+H_2\)
Theo PT: \(n_{Mg}=n_{H_2}=\dfrac{1}{2}n_{CH_3COOH}=0,1\left(mol\right)\)
\(\Rightarrow m=m_{Mg}=0,1.24=2,4\left(g\right)\)
\(V=V_{H_2}=0,1.22,4=2,24\left(l\right)\)
b, \(C_2H_5OH+O_2\underrightarrow{^{mengiam}}CH_3COOH+H_2O\)
Theo PT: \(n_{C_2H_5OH}=n_{CH_3COOH}=0,2\left(mol\right)\)
\(\Rightarrow m_{C_2H_5OH}=0,2.46=9,2\left(g\right)\)
\(\Rightarrow V_{ddC_2H_5OH}=\dfrac{9,2}{0,8}=11,5\left(ml\right)\)