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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: \(n_{C_2H_5OH}=n_{CH_3COOH}=0,2\left(mol\right)\)
\(\Rightarrow m_{C_2H_5OH}=0,2.46=9,2\left(g\right)\)
\(\Rightarrow V_{ddC_2H_5OH}=\dfrac{9,2}{0,8}=11,5\left(ml\right)\)
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: \(n_{C_2H_5OH}=n_{CH_3COOH}=0,2\left(mol\right)\)
\(\Rightarrow m_{ddC_2H_5OH}=\dfrac{0,2\cdot46}{8\%}=115\left(g\right)\) \(\Rightarrow V_{C_2H_5OH}=\dfrac{115}{0,8}=143,75\left(ml\right)\)
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
\(n_{HCl}=0.2\cdot22.4=4.48\left(mol\right)\)
\(\Leftrightarrow n_{H_2}=2.24\left(mol\right)\)
\(\Leftrightarrow m_{H_2}=n_{H_2}\cdot M=2.24\cdot1=2.24\left(g\right)\)
\(n_{MgCl_2}=2.24\left(mol\right)\)
\(\Leftrightarrow n_{Mg}=2.24\left(mol\right)\)
\(\Leftrightarrow m_{Mg}=2.24\cdot24=53.76\left(g\right)\)
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: nN2 = 0,22 (mol)
⇒ nNO3- = 10nN2 = 2,2 (mol)
⇒ m muối = mX + mNO3- = 31 + 2,2.62 = 167,4 (g)
nHNO3 = 12nN2 = 2,64 (mol)
\(\Rightarrow V_{HNO_3}=\dfrac{2,64}{2}=1,32\left(l\right)=1320\left(ml\right)\)
a, \(n_{Fe}=\dfrac{16,8}{56}=0,3\left(mol\right)\)
BT e, có: 3nFe = 2nSO2 ⇒ nSO2 = 0,45 (mol)
⇒ VSO2 = 0,45.22,4 = 10,08 (l)
b, BTNT Fe, có: nFe(OH)3 = nFe = 0,3 (mol)
⇒ mFe(OH)3 = 0,3.107 = 32,1 (g)
\(a,n_{CH_3COOH}=\dfrac{120.20}{100}=24\left(g\right)\\ \rightarrow n_{CH_3COOH}=\dfrac{24}{60}=0,4\left(mol\right)\)
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
\(\rightarrow V_{ddC_2H_5OH}=\dfrac{0,4.46.100}{0,8.46}=50\left(ml\right)\)
nH2 = 85,12 : 22,4 = 3,8 (mol) ; nH2O = VH2O.D = 108 (g) => nH2O = 108/18 = 6 (mol)
PTHH:
2Na + 2C2H5OH → 2C2H5ONa + H2↑
x → 0,5x (mol)
2Na + 2H2O → 2NaOH + H2↑
6 → 3 (mol)
Ta có: nH2 = 0,5x + 3 = 3,8
=> x = 1,6 (mol) = nC2H5OH
mC2H5OH = 1,6.46 = 73,6 (g)
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
nCH3COOH= 0.2 mol
Mg + 2CH3COOH --> (CH3COO)2Mg + H2
Từ PTHH:
nMg= 0.1 mol
mMg=2.4g
nH2= 0.1 mol
VH2= 2.24l
C2H5OH + O2 -mg-> CH3COOH + H2O
0.2___________________0.2
mC2H5OH= 9.2g
VC2H5OH=9.2/0.8=11.5ml
V hhr= 11.5*100/8=143.75ml