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![](https://rs.olm.vn/images/avt/0.png?1311)
1)
$n_{Na_2O} = \dfrac{6,2}{62} = 0,1(mol)$
$Na_2O + H_2O \to 2NaOH$
$n_{NaOH} = 2n_{Na_2O} = 0,2(mol)$
$m_{dd} = 6,2 + 193,8 = 200(gam) \Rightarrow C\%_{NaOH} = \dfrac{0,2.40}{200}.100\% = 4\%$
2)
$n_{K_2O} = \dfrac{23,5}{94} = 0,25(mol)$
$K_2O + H_2O \to 2KOH$
$n_{KOH} = 2n_{K_2O} = 0,5(mol) \Rightarrow C_{M_{KOH}} = \dfrac{0,5}{0,5} = 1M$
3) $n_{Na_2O} = \dfrac{12,4}{62} = 0,2(mol)$
$Na_2O + H_2O \to 2NaOH$
$n_{NaOH} = 2n_{Na_2O} = 0,4(mol)$
$C_{M_{NaOH}} = \dfrac{0,4}{0,5} =0,8M$
4)
$Na_2SO_3 + 2HCl \to 2NaCl +S O_2 + H_2O$
Theo PTHH :
$n_{SO_2} = n_{Na_2SO_3} = \dfrac{12,6}{126} = 0,1(mol)$
$V_{SO_2} = 0,1.22,4 = 2,24(lít)$
5) $n_{CaO} = \dfrac{5,6}{56} = 0,1(mol)$
$CaO + 2HCl \to CaCl_2 + H_2O$
Theo PTHH :
$n_{HCl} = 2n_{CaO} = 0,2(mol) \Rightarrow m_{dd\ HCl} = \dfrac{0,2.36,5}{14,6\%} = 50(gam)$
![](https://rs.olm.vn/images/avt/0.png?1311)
\(n_{K2O}=\dfrac{23,5}{94}=0,25\left(mol\right)\)
Pt : \(K_2O+H_2O\rightarrow2KOH|\)
1 1 2
0,25 0,5
\(n_{KOH}=\dfrac{0,25.2}{1}=0,5\left(mol\right)\)
\(C_{M_{ddKOH}}=\dfrac{0,5}{0,5}=1\left(M\right)\)
⇒ Chọn câu : A
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![](https://rs.olm.vn/images/avt/0.png?1311)
\(n_{K_2O}=\dfrac{23,5}{94}=0,25\left(mol\right)\)
PT: \(K_2O+H_2O\rightarrow2KOH\)
Theo PT: \(n_{KOH}=2n_{K_2O}=0,5\left(mol\right)\)
\(\Rightarrow C_{M_{KOH}}=\dfrac{0,5}{0,5}=1\left(M\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
mH2O=71,8.1=71,8(g)
nK2O=28,2/94=0,3(mol)
PTHH: K2O + H2O -> 2 KOH
0,3____________0,6(mol)
mKOH= 0,6.56=33,6(g)
mddKOH= mK2O + mH2O= 28,2+71,8=100(g)
=>C%ddB=C%ddKOH=(33,6/100).100=33,6%
![](https://rs.olm.vn/images/avt/0.png?1311)
K + H2O -------> KOH + 1/2 H2
nK = 5,85/39=0,15 (mol)
Theo PT : nKOH=nK = 0,15 (mol)
=> CM KOH = n/V = 0,15/0,1=1,5M
=> Chọn C
Số mol của kali
nK = \(\dfrac{m_K}{M_K}=\dfrac{5,85}{39}=0,15\left(mol\right)\)
Pt : 2K + 2H2O → 2KOH + H2\(|\)
2 2 2 1
0,15 0,15
Số mol của dung dịch kali hidroxit
nKOH= \(\dfrac{0,15.2}{2}=0,15\left(mol\right)\)
Nồng độ mol của dung dịch kali hidroxit
CMKOH = \(\dfrac{0,15}{0,1}=1,5\left(M\right)\)
⇒ Chọn câu : C
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![](https://rs.olm.vn/images/avt/0.png?1311)
Câu 1:
\(n_{K2O}=\frac{9,4}{39.2+16}=0,1\left(mol\right)\)
\(K_2O+H_2O\rightarrow2KOH\)
0,1_____________0,2
\(C\%_{KOH}=\frac{0,2.\left(39+17\right)}{150,6+9,4}.100\%=7\%\)
\(KOH+HCl\rightarrow KCl+H_2O\)
0,2______0,2__________________
\(\Rightarrow V_{dd_{HCl}}=\frac{0,2}{0,5}=0,5\left(l\right)\)
Câu 2:
a, \(n_{K2O}=\frac{23,5}{39.2+16}=0,25\left(mol\right)\)
\(2n_{K2O}=n_{KOH}\Rightarrow n_{KOH}=0,25.2=0,5\left(mol\right)\)
\(C\%_{KOH}=\frac{0,5.\left(39+17\right)}{176,5+23,5}.100\%=14\%\)
b, \(n_{KOH}=2n_{K2SO4}\Rightarrow n_{K2SO4}=\frac{0,5}{2}=0,25\)
\(n_{H2SO4}=n_{K2SO4}=0,25\)
\(m_{dd_{H2SO4}}=\frac{0,25.98}{20\%}=122,5\left(g\right)\)
c,
mdd sau phản ứng=mddA+mddH2SO4
m dd sau phản ứng \(=23,5+176,5+122,5=322,5\)
\(C\%_{K2SO4}=\frac{0,25.\left(39.2+32+16.4\right)}{322,5}.100\%=13,49\%\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(n_{K2O}=\dfrac{9,4}{94}=0,1\left(mol\right)\)
Pt : \(K_2O+H_2O\rightarrow2KOH|\)
1 1 2
0,1 0,2
a) \(n_{KOH}=\dfrac{0,1.2}{1}=0,2\left(mol\right)\)
200ml = 0,2l
\(C_{M_{ddKOH}}=\dfrac{0,2}{0,2}=1\left(M\right)\)
b) Pt : \(KOH+HCl\rightarrow KCl+H_2O|\)
1 1 1 1
0,1 0,1
\(n_{HCl}=\dfrac{0,1.1}{1}=0,1\left(mol\right)\)
\(m_{HCl}=0,1.36,5=3,65\left(g\right)\)
\(m_{ddHCl}=\dfrac{3,65.100}{10}=36,5\left(g\right)\)
c) \(CO_2+2KOH\rightarrow K_2CO_3+H_2O|\)
1 2 1 1
0,05 0,1
\(n_{CO2}=\dfrac{0,1.1}{2}=0,05\left(mol\right)\)
\(V_{CO2\left(dktc\right)}=0,05.22,4=1,12\left(l\right)\)
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\(n_{K2O}=\dfrac{23,5}{94}=0,25\left(mol\right)\)
Pt : \(K_2O+H_2O\rightarrow2KOH|\)
1 1 2
0,25 0,5
\(n_{KOH}=\dfrac{0,25.2}{1}=0,5\left(mol\right)\)
\(C_{M_{ddKOH}}=\dfrac{0,5}{0,5}=1\left(M\right)\)
⇒ Chọn câu : B
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