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\(a,n_{Na_2SO_3}=\dfrac{12,6}{126}=0,1mol\\ Na_2SO_3+2HCl\rightarrow2NaCl+SO_2+H_2O\\ n_{SO_2}=n_{Na_2SO_4}=0,1mol\\ V_{SO_2}=0,1.22,4=2,24l\\ b,n_{HCl}=0,1.2=0,2mol\\ C_{M_{HCl}}=\dfrac{0,2}{0,15}=\dfrac{4}{3}M\\ c,n_{NaOH}=\dfrac{40.10}{100.40}=0,1mol\\ T=\dfrac{0,1}{0,1}=1\\ \Rightarrow Tạo,NaHSO_3\\ NaOH+SO_2\rightarrow NaHSO_3\\ m_{NaHSO_3}=0,1.64+0,1.40=10,4g\)
a) \(n_{CO_2}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\)
PTHH: CO2 + 2NaOH → Na2CO3 + H2O
Mol: 0,15 0,3 0,15
\(C_{M_{ddNaOH}}=\dfrac{0,3}{0,2}=1,5M\)
b) Na2CO3: natri cacbonat
\(m_{Na_2CO_3}=0,15.106=15,9\left(g\right)\)
c)
PTHH: 2NaOH + H2SO4 → Na2SO4 + 2H2O
Mol: 0,15 0,075
\(V_{ddH_2SO_4}=\dfrac{0,075}{1}=0,075\left(l\right)=75\left(ml\right)\)
a) CO2+ 2NaOH→ Na2CO3+ H2O
(mol) 0,05 0,1 0,05 0,05
b) \(n_{CO_2}=\dfrac{V}{22,4}=\dfrac{1,12}{22,4}=0,05\left(mol\right)\)
\(m_{Na_2CO_3}=n.M=0,05.106=5,3\)(g)
c)đổi: 200ml=0,2 lít
\(C_{M_{NaOH}}=\dfrac{n}{V}=\dfrac{0,1}{0,2}=0,5M\)
a, \(H_2SO_4+2NaOH\rightarrow Na_2SO_4+2H_2O\)
b, \(n_{H_2SO_4}=0,2.1,5=0,3\left(mol\right)\)
Theo PT: nNa2SO4 = nH2SO4 = 0,3 (mol) ⇒ mNa2SO4 = 0,3.142 = 42,6 (g)
nNaOH = 2nH2SO4 = 0,6 (mol) ⇒ mNaOH = 0,6.40 = 24 (g)
c, \(n_{CO_2}=\dfrac{8,96}{22,4}=0,4\left(mol\right)\)
\(\Rightarrow\dfrac{n_{NaOH}}{n_{CO_2}}=\dfrac{0,6}{0,4}=1,5\) → Pư tạo NaHCO3 và Na2CO3.
PT: \(2NaOH+CO_2\rightarrow Na_2CO_3+H_2O\)
\(NaOH+CO_2\rightarrow NaHCO_3\)
\(n_{Na_2SO_3}=\dfrac{12,6}{126}=0,1mol\)
\(Na_2SO_3+H_2SO_4\rightarrow Na_2SO_4+SO_2+H_2O\)
a) \(n_{SO_2}=n_{Na_2SO_3}=0,1mol\) \(\Rightarrow V=2,24l\)
b) \(n_{H_2SO_4}=n_{Na_2SO_3}=0,1mol\) \(\Rightarrow C_M=\dfrac{0,1}{0,2}=0,5M\)
c) \(m_{Na_2SO_4}=0,1\cdot142=14,2g\)
a, \(HCl+NaOH\rightarrow NaCl+H_2O\)
b, \(n_{HCl}=0,3.2=0,6\left(mol\right)\)
\(n_{NaOH}=n_{NaCl}=n_{HCl}=0,6\left(mol\right)\)
\(\Rightarrow m_{ddNaOH}=\dfrac{0,6.40}{20\%}=120\left(g\right)\)
c, \(m_{NaCl}=0,6.58,5=35,1\left(g\right)\)
\(m_{NaOH}=\dfrac{200\cdot8}{100}=16\left(g\right)\Rightarrow n_{NaOH}=\dfrac{16}{40}=0,4mol\)
\(NaOH+HCl\rightarrow NaCl+H_2O\)
0,4 0,4 0,4 0,4
a)\(m_{HCl}=0,4\cdot36,5=14,6\left(g\right)\)
\(\Rightarrow m_{ddHCl}=\dfrac{14,6}{7,3}\cdot100=200\left(g\right)\)
b)\(m_{NaCl}=0,4\cdot58,5=23,4\left(g\right)\)
\(m_{H_2O}=0,4\cdot18=7,2\left(g\right)\)
\(m_{ddsau}=200+200-7,2=392,8\left(g\right)\)
\(\Rightarrow C\%=\dfrac{23,4}{392,8}\cdot100=5,96\%\)
c) \(n_{SO_2}=\dfrac{6,72}{22,4}=0,3mol\)
\(2NaOH+SO_2\rightarrow Na_2SO_4+H_2O\)
0,4 0,3 0,3 0,3
\(m_{Na_2SO_4}=0,3\cdot142=42,6\left(g\right)\)
\(a,PTHH:SO_2+Ca(OH)_2\to CaSO_3\downarrow+H_2O\\ b,n_{Ca(OH)_2}=0,7.0,01=0,007(mol)\\ n_{SO_2}=\dfrac{0,112}{22,4}=0,005(mol)\)
Vì \(\dfrac{n_{SO_2}}{1}<\dfrac{n_{Ca(OH)_2}}{1}\) nên \(Ca(OH)_2\) dư
\(\Rightarrow n_{CaSO_3}=n_{H_2O}=0,005(mol)\\ \Rightarrow m_{CaSO_3}=0,005.120=0,6(g)\\ m_{H_2O}=0,005.18=0,09(g)\)
PT: \(CH_3COOH+NaOH\rightarrow CH_3COONa+H_2O\)
Ta có: \(n_{CH_3COONa}=\dfrac{9,84}{82}=0,12\left(mol\right)\)
Theo PT: \(n_{CH_3COOH}=n_{NaOH}=n_{CH_3COONa}=0,12\left(mol\right)\)
\(\Rightarrow V_{ddCH_3COOH}=\dfrac{0,12}{0,5}=0,24\left(l\right)\)
\(m_{NaOH}=0,12.40=4,8\left(g\right)\Rightarrow m_{ddNaOH}=\dfrac{4,8}{20\%}=24\left(g\right)\)
\(2NaOH+SO_2-->Na_2SO_3+H_2O\left(1\right)\)
0,2________0,1__________0,1
\(n_{SO_2}=0,1\left(mol\right)\)
a) \(m_{Na_2SO_3}=0,1.126=12,6\left(g\right)\)
b) \(m_{d^2_{NaOH}}=\frac{0,2.40.100}{25}=32\left(g\right)\)
c) \(2NaOH+H_2SO_4-->Na_2SO_4+2H_2O\left(2\right)\)
0,2___________0,1
=> \(V_{d^2H_2SO_4}=\frac{0,1.98.100}{20.1,14}=42,98\left(ml\right)\)