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a, \(m_{CH_3COOH}=100.6\%=6\left(g\right)\Rightarrow n_{CH_3COOH}=\dfrac{6}{60}=0,1\left(mol\right)\)
PT: \(CH_3COOH+NaHCO_3\rightarrow CH_3COONa+CO_2+H_2O\)
Theo PT: \(n_{NaHCO_3}=n_{CH_3COONa}=n_{CO_2}=n_{CH_3COOH}=0,1\left(mol\right)\)
\(\Rightarrow m_{NaHCO_3}=0,1.84=8,4\left(g\right)\)
\(V_{CO_2}=0,1.22,4=2,24\left(l\right)\)
b, Ta có: m dd sau pư = 100 + 8,4 - 0,1.44 = 104 (g)
\(\Rightarrow C\%_{CH_3COONa}=\dfrac{0,1.82}{104}.100\%\approx7,88\%\)
Ta có: \(m_{CH_3COOH}=100.12\%=12\left(g\right)\Rightarrow n_{CH_3COOH}=\dfrac{12}{60}=0,2\left(mol\right)\)
PT: \(CH_3COOH+NaHCO_3\rightarrow CH_3COONa+CO_2+H_2O\)
Theo PT: \(n_{NaHCO_3}=n_{CH_3COONa}=n_{CO_2}=n_{CH_3COOH}=0,2\left(mol\right)\)
\(\Rightarrow m_{ddNaHCO_3}=\dfrac{0,2.84}{8,4\%}=200\left(g\right)\)
Ta có: m dd sau pư = m dd CH3COOH + m dd NaHCO3 - mCO2 = 100 + 200 - 0,2.44 = 291,2 (g)
\(\Rightarrow C\%_{CH_3COONa}=\dfrac{0,2.82}{291,2}.100\%\approx2,82\%\)
\(m_{CH_3COOH}=12\%.100=12\left(g\right)\\ n_{CH_3COOH}=\dfrac{12}{60}=0,2\left(mol\right)\)
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) MgO + 2HCl ---> MgCl2 + H2O (1)
MgCO3 + 2HCl ---> MgCl2 + H2O + CO2 (2)
b) \(n_{CO_2}=\dfrac{m}{M}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\)
=> \(m_{MgCO_3}=n.M=0,1.84=8,4\left(g\right)\)
\(m_{MgO}=16-8,4=8\left(g\right)\)
c) \(n_{MgO}=\dfrac{m}{M}=\dfrac{8}{40}=0,2\left(mol\right)\)
=> \(n_{HCl\left(1\right)}=0,4\left(mol\right)\); nHCl(2) = 0,2(mol)
=> nHCl = 0.4 + 0,2 = 0,6 (mol)
=> VHCl = \(\dfrac{n}{C_M}=\dfrac{0,6}{1,5}=0,4\left(l\right)=400\left(ml\right)\)
\(n_{Fe}=\dfrac{5,6}{56}=0,1\left(mol\right)\)
PT: \(Fe+2CH_3COOH\rightarrow\left(CH_3COO\right)_2Fe+H_2\)
a, Theo PT: \(n_{CH_3COOH}=2n_{Fe}=0,2\left(mol\right)\Rightarrow m_{CH_3COOH}=0,2.60=12\left(g\right)\)
\(\Rightarrow m_{ddCH_3COOH}=\dfrac{12}{10\%}=120\left(g\right)\)
\(n_{H_2}=n_{Fe}=0,1\left(mol\right)\Rightarrow V_{H_2}=0,1.22,4=2,24\left(l\right)\)
b, Theo PT: \(n_{\left(CH_3COO\right)_2Fe}=n_{Fe}=0,1\left(mol\right)\)
Ta có: m dd sau pư = 5,6 + 120 - 0,1.2 = 125,4 (g)
\(\Rightarrow C\%_{\left(CH_3COO\right)_2Fe}=\dfrac{0,1.174}{125,4}.100\%\approx13,88\%\)
a)
$CH_3COOH + NaHCO_3 \to CH_3COONa + CO_2 + H_2O$
b)
n NaHCO3 = n CH3COOH = 100.12%/60 = 0,2(mol)
m dd NaHCO3 = 0,2.84/8% = 210(gam)
c)
n CO2 = n CH3COOH = 0,2(mol)
=> V CO2 = 0,2.22,4 = 4,48(lít)
d)
m dd = m dd CH3COOH + m dd NaHCO3 - m CO2 = 100 + 210 - 0,2.44 = 301,2(gam)
C% CH3COONa = 0,2.82/301,2 .100% = 5,44%
\(n_{FeCl_3}=\dfrac{48,75}{162,5}=0,3(mol)\\ 3NaOH+FeCl_3\to Fe(OH)_3\downarrow+3NaCl\\ \Rightarrow n_{Fe(OH)_3}=0,3(mol);n_{NaOH}=n_{NaCl}=0,9(mol)\\ a,m_{Fe(OH)_3}=0,3.107=32,1(g)\\ b,m_{dd_{NaOH}}=\dfrac{0,9.40}{10\%}=360(g)\\ c,C\%_{NaCl}=\dfrac{0,9.58,5}{360+48,75-32,1}.100\%=13,98\%\\ \)
\(d,2Fe(OH)_3+3H_2SO_4\to Fe_2(SO_4)_3+6H_2O\\ \Rightarrow n_{H_2SO_4}=0,45(mol)\\ \Rightarrow m_{dd_{H_2SO_4}}=\dfrac{0,45.98}{20\%}=220,5(g)\\ \Rightarrow V_{dd_{H_2SO_4}}=\dfrac{220,5}{1,14}=193,42(ml)\)
\(a.n_{CH_3COOH}=\frac{60.5}{100.60}=0,05\left(mol\right)\)
\(PTHH:CH_3COOH+NaHCO_3\rightarrow CH_3COONa+H_2O+CO_2\)
`(mol)`_______`0,05`_______`0,05`_________`0,05`____________`0,05`_
\(m_{ddNaHCO_3}=\frac{0,05.84.100}{8,4}=50\left(g\right)\)
\(b.V_{CO_2}=0,05.22,4=1,12\left(l\right)\)
Bạn xem lại câu `c`, làm gì có hh đâu nhờ =)