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PTHH: \(CH_3COOH+NaOH\rightarrow CH_3COONa+H_2O\)
Ta có: \(n_{NaOH}=\dfrac{30\cdot20\%}{40}=0,15\left(mol\right)=n_{CH_3COONa}=n_{CH_3COOH}\)
\(\Rightarrow\left\{{}\begin{matrix}m_{CH_3COONa}=0,15\cdot82=12,3\left(g\right)\\C_{M_{CH_3COOH}}=\dfrac{0,15}{0,5}=0,3\left(M\right)\end{matrix}\right.\)
\(CH_3COOH+NaCl\rightarrow CH_3COONa+HCl\)
\(2CH_3COOH+CaCO_3\rightarrow\left(CH_3COO\right)_2Ca+H_2O+CO_2\)
2 1 1 1 1 (mol)
0,08 0,04 0,04 0,04 0,04 (mol)
\(nCO_2=\dfrac{0,896}{22,4}=0,04\left(mol\right)\)
\(mCaCO_3=0,04.100=4\left(g\right)\)
=> \(mNaCl=12,5-4=8,5\left(g\right)\)
( không thấy hh B )
c ) .
\(CO_2+2NaOH\rightarrow Na_2CO_3+H_2O\)
1 2 1 1 (mol)
0,04 0,08 0,04 0,04 (mol)
\(mNa_2CO_3=0,04.106=4,24\left(g\right)\)
\(mNa_2CO_{3\left(thựctế\right)}=\)\(\dfrac{4,24.85\%}{100\%}=3,604\left(g\right)\)
Bài 1
\(a,n_{CuO}=\dfrac{16}{80}=0,2\left(mol\right)\\ CuO+2HCl\xrightarrow[]{}CuCl_2+H_2O\\ n_{CuCl_2}=n_{CuO}=0,2mol\\ m_{CuCl_2}=0,2.135=27\left(g\right)\\ b.n_{HCl}=0,2.2=0,4\left(mol\right)\\ C_{MHCl}=\dfrac{0,4}{0,5}=0,8\left(M\right)\)
Bài 5
\(a,n_{NaOH}=0,2.1=0,2\left(mol\right)\\ 2NaOH+H_2SO_4\xrightarrow[]{}Na_2SO_4+2H_2O\\ n_{H_2SO_4}=0,2:2=0,1\left(mol\right)\\ C_{MH_2SO_4}=\dfrac{0,1}{0,4}=0,25\left(M\right)\\ b,n_{Na_2SO_4}=0,2:2=0,1\left(mol\right)\\ C_{MNa_2SO_4}=\dfrac{0,1}{0,2+0,4}=\dfrac{1}{6}\left(M\right)\\ c,m_{Na_2SO_4}=0,1.142=14,2\left(g\right)\)
\(n_{FeCl_3}=\dfrac{48,75}{162,5}=0,3(mol)\\ 3NaOH+FeCl_3\to Fe(OH)_3\downarrow+3NaCl\\ \Rightarrow n_{Fe(OH)_3}=0,3(mol);n_{NaOH}=n_{NaCl}=0,9(mol)\\ a,m_{Fe(OH)_3}=0,3.107=32,1(g)\\ b,m_{dd_{NaOH}}=\dfrac{0,9.40}{10\%}=360(g)\\ c,C\%_{NaCl}=\dfrac{0,9.58,5}{360+48,75-32,1}.100\%=13,98\%\\ \)
\(d,2Fe(OH)_3+3H_2SO_4\to Fe_2(SO_4)_3+6H_2O\\ \Rightarrow n_{H_2SO_4}=0,45(mol)\\ \Rightarrow m_{dd_{H_2SO_4}}=\dfrac{0,45.98}{20\%}=220,5(g)\\ \Rightarrow V_{dd_{H_2SO_4}}=\dfrac{220,5}{1,14}=193,42(ml)\)
a) \(n_{CH_3COOH}=\dfrac{120.20\%}{60}=0,4\left(mol\right)\)
\(n_{Na_2CO_3}=\dfrac{53.30\%}{106}=0,15\left(mol\right)\)
PTHH: Na2CO3 + 2CH3COOH --> 2CH3COONa + CO2 + H2O
Xét tỉ lệ: \(\dfrac{0,15}{1}< \dfrac{0,4}{2}\) => Na2CO3 hết, CH3COOH dư
PTHH: Na2CO3 + 2CH3COOH --> 2CH3COONa + CO2 + H2O
0,15-------->0,3-------------->0,3------->0,15
=> \(m_{CH_3COONa}=0,3.82=24,6\left(g\right)\)
b) mdd sau pư = 120 + 53 - 0,15.44 = 166,4 (g)
=> \(C\%=\dfrac{24,6}{166,4}.100\%=14,78\text{%}\)
PTHH: CH3COOH + NaHCO3 ➞ CH3COONa + H2O + CO2
Ta có: mNAHCO3=(200.84)/100= 16.8 gam
nNAHCO3= 16.8/84= 0.2 mol
mCH3COOH= 0.2*60= 12 gam
Câu a) mddCH3COOH= (12*100)/6= 200 gam
Câu b) mddCH3COONA= mddCH3COOH + mdd NAHCO3= 200+200=400 gam
mCH3COONA= 0.2*82=16.4 gam
C%dd CH3COONA= (16.4*100)/400= 4.1%
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