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PTHH: \(CO_2+2NaOH\rightarrow Na_2CO_3+H_2O\)
Ta có: \(n_{CO_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\)
\(\Rightarrow\left\{{}\begin{matrix}n_{Na_2CO_3}=0,3\left(mol\right)\\n_{NaOH}=0,6\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}C\%_{NaOH}=\dfrac{0,6\cdot40}{240}\cdot100\%=10\%\\C\%_{Na_2CO_3}=\dfrac{0,3\cdot106}{240+0,3\cdot44}\cdot100\%\approx12,56\%\end{matrix}\right.\)
Na2O + H2O -> 2NaOH (1)
nNa2O=0,25(mol)
Theo PTHH 1 ta có:
nNaOH=2nNa2O=0,5(mol)
CM dd NaOH=\(\dfrac{0,5}{0,5}=1M\)
NaOH + HCl -> NaCl + H2O
Theo PTHH 2 ta có:
nNaOH=nHCl=0,5(mol)
CM dd HCl=\(\dfrac{0,5}{0,5}=1M\)
\(a,PTHH:Na_2O+H_2O\rightarrow2NaOH\\ \Rightarrow n_{NaOH}=2n_{Na_2O}=2\cdot\dfrac{37,2}{62}=0,6\cdot2=1,2\left(mol\right)\\ \Rightarrow C_{M_{NaOH}}=\dfrac{1,2}{0,5}=2,4M\\ b,PTHH:2NaOH+H_2SO_4\rightarrow Na_2SO_4+H_2O\\ \Rightarrow n_{H_2SO_4}=\dfrac{1}{2}n_{NaOH}=0,6\left(mol\right)\\ \Rightarrow m_{H_2SO_4}=0,6\cdot98=58,8\left(g\right)\\ \Rightarrow m_{dd_{H_2SO_4}}=\dfrac{58,8\cdot100\%}{20\%}=294\left(g\right)\\ \Rightarrow V_{dd}=\dfrac{294}{1,14}\approx257,9\left(ml\right)\)
Anh bổ sung câu c)
\(C_{MddNa_2SO_4}=\dfrac{0,25}{0,09879+0,5}=0,4175\left(M\right)\)
nNa2O = 0,125 mol
a. Na2O + H2O --------> NaOH
0,125 mol ----------------> 0,125 mol
--> CM(NaOH) n/V = 0,125/ 0,25 = 0,5 M
b. H2SO4 + 2NaOH ------> Na2SO4 + H2O
....0,0625 <---0,125 mol
--> mH2SO4(nguyên chất) = 0,0625*98 = 6,125 g
--> mH2SO4(20%) = 6,125/20% = 30,625 g
suy ra V = m/D = 30,625 / 1,14 = 26,86 ml
nNa2O = 0,125 mol
a. Na2O + H2O --------> NaOH
0,125 mol ----------------> 0,125 mol
--> CM(NaOH) n/V = 0,125/ 0,25 = 0,5 M
b. H2SO4 + 2NaOH ------> Na2SO4 + H2O
....0,0625 <---0,125 mol
--> mH2SO4(nguyên chất) = 0,0625*98 = 6,125 g
--> mH2SO4(20%) = 6,125/20% = 30,625 g
suy ra V = m/D = 30,625 / 1,14 = 26,86 ml
Na2O + H2O → 2NaOH
1 1 2
0,1 0,2
a). nNa2O=\(\dfrac{6,2}{62}\)= 0,1(mol)
CM=\(\dfrac{n}{V}\)=\(\dfrac{0,1}{4}\)= 0,025M
b). Na2O + H2SO4 → Na2SO4 + H2O
1 1 1 1
0,1 0,1
mH2SO4= n.M = 0,1 . 98 = 9,8g
⇒mddH2SO4= mct=\(\dfrac{mct.100\%}{C\%}\)= \(\dfrac{9,8.100}{20}\)= 49(g).
\(n_{Na2O}=\frac{15,5}{62}=0,25\left(mol\right)\)
a) \(Na_2O+H_2O\rightarrow2NaOH\)
__0,25______________0,5_______ (mol)
\(CM_{NaOH}=\frac{0,5}{0,5}=1M\)
b) \(2NaOH+H_2SO_4\rightarrow Na_2SO_4+2H_2O\)
__0,5 _______0,25_______0,25_______ 0,5_______ (mol)
\(m_{H2SO4}=0,25.98=24,5\left(g\right)\)
\(m_{dd_{H2SO4}}=\frac{24,5.100\%}{20\%}=122,5\left(g\right)\)
\(\Rightarrow V_{dd_{H2SO4}}=\frac{122,5}{1,14}=107,456\left(ml\right)\approx0,107\left(l\right)\)
c) \(V_{dd_{mơi}}=0,5+0,107=0,607\left(l\right)\)
\(\Rightarrow CM_{Na2SO4}=\frac{0,25}{0,607}=0,412M\)
\(\Rightarrow CM_{H2O}=\frac{0,5}{0,607}=0,824M\)
\(n_{K_2O}=\dfrac{1,88}{94}=0,02(mol)\\ a,K_2O+H_2O\to 2KOH\\ b,n_{KOH}=0,04(mol)\\ \Rightarrow C_{M_{KOH}}=\dfrac{0,04}{0,5}=0,08M\\ c,n_{KOH}=0,04.50\%=0,02(mol)\\ KOH+HCl\to KCl+H_2O\\ \Rightarrow n_{HCl}=0,02(mol)\\ \Rightarrow m_{dd_{HCl}}=\dfrac{0,02.36,5}{7,3\%}=10(g)\)
1) K2CO3 +2HCl ------> 2KCl + CO2 +H2O
a) nHCl=\(\dfrac{200\cdot7,3\%}{36,5}\)=0,4 mol
Theo pthh nCO2= nK2CO3=1/2nHCl=0,2 mol
=>VCO2=0.2*22.4=4,48 (lít)
b)mdd K2CO3=\(\dfrac{0,2\cdot138}{20\%}\)=138(g)
c)mddA =mdd HCl +mdd K2CO3-mCO2=200+138-0.2*2=337,6
Theo pt nKCl=nHCl=0,4 mol
=>C%=\(\dfrac{0,4\cdot74,5}{337,6}\)=8,83%
2) 2Al +6HCl------> 2AlCl3 +3H2
a)nAl=0,2 mol
nHCl=\(\dfrac{365\cdot12\%}{36,5}\)=1,2 mol
Xét tỉ lệ \(\dfrac{n_{Al}}{2}\) < \(\dfrac{n_{HCl}}{6}\) => nAl hết nHCl dư, tính theo Al
Theo pt nAlCl3=nAl=0,2 mol
nH2=3/2nAl=0,3 mol
=>mAlCl3=133.5*0,2=26,7(g)
=>VH2=0,3*22,4=6,72 (l)
b)Theopt nHCl=3nAl=0,6 mol
mddA=mAl+mddHCl-mH2=5,4+365-0,3*2=369,8 (g)
=>C% HCl=\(\dfrac{\left(1,2-0,6\right)\cdot36,5\cdot100}{369,8}\)=5,92%
=>C%AlCl3=\(\dfrac{26,7\cdot100}{369,8}\)=7,22%
c) HCl+ KOH ---------> KCl+ H2O
Ta có nHCl dư=0,6 mol
Theo pt nKOH=nHCl=0,6
=>V=\(\dfrac{0,6}{1,5}\)=0,4 (l)