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Gọi $n_{Na_2O} = 2a(mol) \Rightarrow n_{K_2O} = a(mol)$
$\Rightarrow 2a.62 + 94a = 21,8 \Rightarrow a = 0,1(mol)$
$Na_2O + H_2O \to 2NaOH$
$K_2O + H_2O \to 2KOH$
$n_{NaOH} = 2n_{Na_2O} = 0,4(mol)$
$n_{KOH} = 2n_{K_2O} = 0,2(mol)$
$C_{M_{NaOH}} = \dfrac{0,4}{0,5} = 0,8M$
$C_{M_{KOH}} = \dfrac{0,2}{0,5} = 0,4M$
$m_{dd} = D.V = 1,04.500 = 520(gam)$
$C\%_{NaOH} = \dfrac{0,4.40}{520}.100\% = 3,1\%$
$C\%_{KOH} = \dfrac{0,2.56}{520}.100\% = 2,15\%$
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 :
$n_{HCl} = 2n_{CaO} = 0,2(mol) \Rightarrow m_{dd\ HCl} = \dfrac{0,2.36,5}{14,6\%} = 50(gam)$
\(n_{Na_2O}=\dfrac{6,2}{62}=0,1\left(mol\right)\\ Na_2O+H_2O\rightarrow2NaOH\\ C_{MddA}=C_{MddNaOH}=\dfrac{0,2}{0,4}=0,5\left(M\right)\)
\(n_{Na_2O}=\dfrac{12,4}{62}=0,2mol\)
\(Na_2O+H_2O\rightarrow2NaOH\)
0,2 \(\rightarrow\) 0,2 \(\rightarrow\) 0,4
\(C_{M_{NaOH}}=\dfrac{0,4}{\dfrac{500}{1000}}=0,8M\)
a)
$Na_2O + H_2O \to 2NaOH$
n Na2O = 15,5/62 = 0,25(mol)
n NaOH = 2n Na2O = 0,5(mol)
=> CM NaOH = 0,5/0,5 = 1M
b) $2NaOH + H_2SO_4 \to Na_2SO_4 + 2H_2O$
n H2SO4 = 1/2 n NaOH = 0,25(mol)
=> m dd H2SO4 = 0,25.98/20% = 122,5(gam)
=> V dd H2SO4 = m / D = 122,5/1,14 =107,46(ml)
c) n Na2SO4 = n H2SO4 = 0,25(mol)
CM Na2SO4 = 0,25/0,10746 = 2,33M
a)
`\(Na_2O++H_{2_{ }}O->2NaOH\)
\(n_{Na_2O}=\dfrac{15,5}{62}=0,25mol\)
\(n_{Na_2O}=2n_{Na_2O}=0,5mol\)
\(C_{M_{NaOH}}=\dfrac{0,5}{0,5}\)=1M
\(n_{K2O}=\dfrac{9,4}{94}=0,1\left(mol\right)\)
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
\(n_{CO2}=\dfrac{0,1.1}{2}=0,05\left(mol\right)\)
\(V_{CO2\left(dktc\right)}=0,05.22,4=1,12\left(l\right)\)
Chúc bạn học tốt
1)
a,\(n_{NaOH}=\dfrac{8}{40}=0,2\left(mol\right)\)
\(C_{M_{ddNaOH}}=\dfrac{0,2}{0,242}=0,83M\)
\(C\%_{ddNaOH}=\dfrac{8.100\%}{242}=3,3\%\)
b,\(n_{H_2SO_4}=0,1.0,15=0,015\left(mol\right)\)
PTHH: 2NaOH + H2SO4 → Na2SO4 + 2H2O
Mol: 0,03 0,015
\(C_{M_{ddNaOH}}=\dfrac{0,03}{0,2}=0,15M\)
Đổi 40 ml = 0,04 l
m H2O = D.V = 40 .1= 40g
\(\rightarrow\) m dd = 3,1+ 40 = 43,1g
Ta có PTHH Na2O + H20 \(\rightarrow\) 2NaOH
nNa2O=\(\frac{3,1}{62}\)= 0.05 mol
PT \(\rightarrow\) 2nNa2O = NaOH = 0,05 .2 =0,1
\(\rightarrow\)CM NaOH = \(\frac{0,1}{0,04}\)= 2,5
C% =\(\frac{0,1.40}{43,1}\) = 9,28%