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a, \(Na_2O+H_2O\rightarrow2NaOH\)
Ta có: \(n_{Na_2O}=\dfrac{15,5}{62}=0,25\left(mol\right)\)
Theo PT: \(n_{NaOH}=2n_{Na_2O}=0,5\left(mol\right)\)
\(\Rightarrow CM_{NaOH}=\dfrac{0,5}{0,5}=1\left(M\right)\)
b, \(2NaOH+H_2SO_4\rightarrow Na_2SO_4+2H_2O\)
Theo PT: \(n_{H_2SO_4}=\dfrac{1}{2}n_{NaOH}=0,25\left(mol\right)\)
\(\Rightarrow m_{ddH_2SO_4}=\dfrac{0,25.98}{20\%}=122,5\left(g\right)\)
\(\Rightarrow V_{ddH_2SO_4}=\dfrac{122,5}{1,14}\approx107,46\left(ml\right)\)
a)
\(n_{Na_2O}=\dfrac{31}{62}=0,5\left(mol\right)\)
PTHH: Na2O + H2O --> 2NaOH
______0,5--------------->1
=> \(C_{M\left(NaOH\right)}=\dfrac{1}{0,5}=2M\)
b)
PTHH: H2SO4 + 2NaOH --> Na2SO4 + 2H2O
______0,5<---------1
=> mH2SO4 = 0,5.98 = 49(g)
=> \(m_{dd\left(H_2SO_4\right)}=\dfrac{49.100}{20}=245\left(g\right)\)
=> \(V_{dd\left(H_2SO_4\right)}=\dfrac{245}{1,14}=214,912\left(ml\right)\)
\(n_{Na_2O}=\dfrac{31}{62}=0,5(mol)\\ a,Na_2O+H_2O\to 2NaOH\\ \Rightarrow n_{NaOH}=1(mol)\\ \Rightarrow C_{M_{NaOH}}=\dfrac{1}{0,5}=2M\\ b,2NaOH+H_2SO_4\to Na_2SO_4+2H_2O\\ \Rightarrow n_{H_2SO_4}=0,5(mol)\\ \Rightarrow m_{dd_{H_2SO_4}}=\dfrac{0,5.98}{20\%}=245(g)\\ \Rightarrow V_{dd_{H_2SO_4}}=\dfrac{245}{1,14}=214,91(ml)\)
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
\(n_{Na_2O}=\dfrac{15.5}{62}=0.25\left(mol\right)\)
\(Na_2O+H_2O\rightarrow2NaOH\)
\(0.25.......................0.5\)
\(C_{M_{NaOH}}=\dfrac{0.5}{0.5}=1\left(M\right)\)
b.
\(2NaOH+H_2SO_4\rightarrow Na_2SO_4+2H_2O\)
\(0.5..............0.25\)
\(V_{dd_{H_2SO_4}}=\dfrac{0.25}{2}=0.125\left(l\right)\)
\(a.n_{Na_2O}=\dfrac{15,5}{62}=0,25\left(mol\right)\\ PTHH:Na_2O+H_2O\rightarrow2NaOH\\ 0,25...................................0,5\left(mol\right)\\ C_{MddNaOH}=\dfrac{0,5}{0,5}=1\left(M\right)\\ b.H_2SO_4+2NaOH\rightarrow Na_2SO_4+2H_2O\\ 0,25............0,5..........0,25\left(mol\right)\\ V_{ddH_2SO_4}=\dfrac{0,25}{2}=0,125\left(l\right)\)
K2O + H2O -> 2KOH (1)
nK2O=\(\dfrac{14,1}{94}=0,15\left(mol\right)\)
Theo PTHH 1 ta có:
2nK2O=nKOH=0,3(mol)
CM dd KOH=\(\dfrac{0,3}{0,5}=0,6M\)
c;
2NaOH + H2SO4 -> Na2SO4 + 2H2O (2)
Theo PTHH 2 ta có:
\(\dfrac{1}{2}\)nNaOH=nH2SO4=0,15(mol)
mH2SO4=98.0,15=14,7(g)
mdd H2SO4=14,7:10%=147(g)
Vdd H2SO4=147:1,14=129(ml)
a, K2O + H2O ->2 KOH
b, nK2O= 0,15 ( mol )
K2O + H2O-> 2KOH
Theo pt 1 1 2 ( mol )
Theo đb 0,15 0,15 0,3 ( mol)
==> CM=\(\dfrac{0.3}{0,5}\) = 0,6 M
c, 2KOH + H2SO4 -> K2SO4 + 2H2O
Theo pt 2 1
Theo đb 0,15 0,075
==> mdd H2S04=\(\dfrac{0,075.98.100\%}{10\%}=73,5\left(g\right)\)
==> Vdd h2so4=\(\dfrac{73,5}{1,14}=64,47\left(ml\right)\)
BaO+H2O -> Ba(OH)2
0,02 0,02
a) CM = n/V = 0,02/0,02 = 1M
b) Ba(OH)2 + H2SO4 -> BaSO4 +2H2O
0,02 0,02
=> m = 0,392 g
D = m/V = 1,14
=> 0,392/V = 1,14 => V = 0,34l
a)
$n_{Na_2O} = \dfrac{15,5}{62} = 0,25(mol)$
$Na_2O + H_2O \to 2NaOH$
$n_{NaOH} = 2n_{Na_2O} = 0,5(mol)$
$C_{M_{NaOH}} = \dfrac{0,5}{0,5} = 1M$
b)
$2NaOH + H_2SO_4 \to Na_2SO_4 + 2H_2O$
$n_{H_2SO_4} = \dfrac{1}{2}n_{NaOH} = 0,25(mol)$
$\Rightarrow m_{dd\ H_2SO_4} = \dfrac{0,25.98}{20\%} = 122,5(gam)$
$\Rightarrow V_{dd\ H_2SO_4} = \dfrac{122,5}{1,14} = 107,46(ml)$
a, \(Na_2O+H_2O\rightarrow2NaOH\)
\(n_{Na_2O}=\dfrac{15,5}{62}=0,25\left(mol\right)\)
\(n_{NaOH}=2n_{Na_2O}=0,5\left(mol\right)\)
\(\Rightarrow C_{M_{NaOH}}=\dfrac{0,5}{0,5}=1\left(M\right)\)
b, \(H_2SO_4+2NaOH\rightarrow Na_2SO_4+2H_2O\)
\(n_{H_2SO_4}=\dfrac{1}{2}n_{NaOH}=0,25\left(mol\right)\)
\(\Rightarrow m_{ddH_2SO_4}=\dfrac{0,25.98}{20\%}=122,5\left(g\right)\)
\(\Rightarrow V_{ddH_2SO_4}=\dfrac{122,5}{1,14}\approx107,46\left(ml\right)\)