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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 )
\(H=\dfrac{0,1}{0,2}.100=50\%\)
Số mol của Mg là 4,8:24=0,2 (mol).
Mg (0,2 mol) + 2CH3COOH (0,4 mol) \(\rightarrow\) Mg(OOCCH3)2 (0,2 mol) + H2.
a) Nồng độ mol cần tìm là 0,4:0,2=2 (mol/l).
b) Khối lượng của magie axetat là 0,2.142=28,4 (g).
Bạn kiểm tra lại giúp mình câu c nhé!
nNaOH = 0,1.2 = 0,2 (mol)
PTHH: CH3COOH + NaOH --> CH3COONa + H2O
0,2<--------0,2
=> \(C_{M\left(dd.CH_3COOH\right)}=\dfrac{0,2}{0,15}=\dfrac{4}{3}M\)
nNaOH = 0,1 . 2 = 0,2 (mol)
PTHH: CH3COOH + NaOH -> CH3COONa + H2O
nCH3COOH = nNaOH = 0,2 (mol)
CM(CH3COOH) = 0,2/0,15 = 1,33M
a) \(n_{\left(CH_3COO\right)_2Mg}=\dfrac{7,1}{142}=0,05\left(mol\right)\)
PTHH: 2CH3COOH + Mg --> (CH3COO)2Mg + H2
0,1<----------------------0,05------->0,05
=> VH2 = 0,05.22,4 = 1,12 (l)
b) \(C_{M\left(dd.CH_3COOH\right)}=\dfrac{0,1}{0,025}=4M\)
a)
\(n_{Na_2CO_3}=\dfrac{10,6}{106}=0,1\left(mol\right)\)
PTHH: Na2CO3 + 2CH3COOH --> 2CH3COONa + CO2 + H2O
0,1--------->0,2------------->0,2------------>0,1
=> mCH3COOH = 0,2.60 = 12 (g)
\(C_{M\left(dd.CH_3COOH\right)}=\dfrac{0,2}{0,4}=0,5M\)
b) \(n_{C_2H_5OH}=\dfrac{13,8}{46}=0,3\left(mol\right)\)
PTHH: CH3COOH + C2H5OH --H2SO4,to--> CH3COOC2H5 + H2O
Xét tỉ lệ: \(\dfrac{0,2}{1}< \dfrac{0,3}{1}\) => Hiệu suất tính theo CH3COOH
\(n_{CH_3COOH\left(pư\right)}=\dfrac{0,2.80}{100}=0,16\left(mol\right)\)
PTHH: CH3COOH + C2H5OH --H2SO4,to--> CH3COOC2H5 + H2O
0,16------------------------------------->0,16
=> \(m_{CH_3COOC_2H_5}=0,16.88=14,08\left(g\right)\)
\(m_{CH_3COOH}=12\%.100=12\left(g\right)\\ n_{CH_3COOH}=\dfrac{12}{60}=0,2\left(mol\right)\)
PTHH: CH3COOH + NaOH ---> CH3COONa + H2O
0,2--------->0,2------------>0,2
\(m_{NaOH}=0,2.40=8\left(g\right)\\ m_{ddNaOH}=\dfrac{8}{8,4\%}=\dfrac{2000}{21}\left(g\right)\\ m_{ddCH_3COONa}=\dfrac{2000}{21}+100=\dfrac{4100}{21}\left(g\right)\\ m_{CH_3COONa}=0,2.82=16,4\left(g\right)\\ C\%_{CH_3COONa}=\dfrac{16,4}{\dfrac{4100}{21}}.100\%=8,4\%\)
`=>` Gợi ý:
`CH3COOH + NaHCO3 => CH3COONa + CO2 + H2O`
`mCH3COOH = 100x12/100 = 12` (g)
`==> nCH3COOH = m/M = 12/60 = 0.2` (mol)
Theo pt: `=> nNaHCO3 = 0.2` (mol)
`==> mNaHCO3 = n.M = 0.2x84 =16.8` (g)
`==> mdd NaHCO3 = 16.8x100/8.4 = 200` (g)
Ta có: `nCH3COONa = 0.2` (mol)
\(a.n_{Mg\left(OH\right)_2}=\dfrac{17,4}{58}=0,3\left(mol\right)\\ Mg\left(OH\right)_2+2HCl\rightarrow MgCl_2+2H_2O\\ n_{HCl}=2n_{Mg\left(OH\right)_2}=0,6\left(mol\right)\\ CM_{HCl}=\dfrac{0,6}{0,2}=3M\\b. n_{Mg\left(OH\right)_2}=n_{MgCl_2}=0,3\left(mol\right)\\ m_{MgCl_2}=0,3.85=25,5\left(g\right)\\c.CM_{MgCl_2}=\dfrac{0,3}{0,2}=1,5M \)
a) \(n_{Mg}=\dfrac{4,2}{24}=0,175\left(mol\right)\)
PTHH: Mg + 2CH3COOH --> (CH3COO)2Mg + H2
0,175----->0,35------------>0,175
=> \(C_{M\left(dd.CH_3COOH\right)}=\dfrac{0,35}{0,1}=3,5M\)
b) \(m_{\left(CH_3COO\right)_2Mg}=0,175.142=24,85\left(g\right)\)
a, \(n_{Mg}=\dfrac{4,8}{24}=0,2\left(mol\right)\)
PT: \(Mg+2CH_3COOH\rightarrow\left(CH_3COO\right)_2Mg+H_2\)
Theo PT: \(n_{CH_3COOH}=2n_{Mg}=0,4\left(mol\right)\)
\(\Rightarrow C_{M_{CH_3COOH}}=\dfrac{0,4}{0,2}=2\left(M\right)\)
b, \(n_{\left(CH_3COO\right)_2Mg}=n_{Mg}=0,2\left(mol\right)\)
\(\Rightarrow m_{\left(CH_3COO\right)_2Mg}=0,2.142=28,4\left(g\right)\)