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a) \(n_{Na_2O}=\dfrac{7,75}{62}=0,125\left(mol\right)\)
PTHH: Na2O + H2O --> 2NaOH
_____0,125------------->0,25
\(C_{M\left(NaOH\right)}=\dfrac{0,25}{0,25}=1M\)
b)
PTHH: 2NaOH + H2SO4 --> Na2SO4 + 2H2O
_______0,25---->0,125
=> mH2SO4 = 0,125.98 = 12,25(g)
=> \(m_{dd}=\dfrac{12,25.100}{20}=61,25\left(g\right)\)
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)
$n_{Na_2O} = \dfrac{15,5}{62} = 0,25(mol)$
$Na_2O + H_2O \to 2NaOH$
$n_{NaOH} = 2n_{Na_2O} = 0,5(mol)$
$C_{M_{NaOH}} = \dfrac{0,5}{0,5} = 1M$
b)
$2NaOH + H_2SO_4 \to Na_2SO_4 + 2H_2O$
$n_{H_2SO_4} = \dfrac{1}{2}n_{NaOH} = 0,25(mol)$
$\Rightarrow m_{dd\ H_2SO_4} = \dfrac{0,25.98}{20\%} = 122,5(gam)$
$\Rightarrow V_{dd\ H_2SO_4} = \dfrac{122,5}{1,14} = 107,46(ml)$
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
\(n_{Na_2O}=\dfrac{6,2}{62}=0,1 \left(mol\right)\)
\(2Na+2H_2O\rightarrow2NaOH+H_2\)
0,1 -----------------> 0,1
\(CM_{base}=CM_{NaOH}=\dfrac{0,1}{0,2}=0,5M\)
b
\(H_2SO_4+2NaOH\rightarrow Na_2SO_4+2H_2O\)
0,05 <------ 0,1
\(V_{H_2SO_4}=\dfrac{0,05}{0,2}=0,25\left(l\right)\Rightarrow V_{dd.H_2SO_4}=\dfrac{0,25.100}{20}=1,25\left(l\right)\)
`n_{Na_2O}={4,6}/{62}\approx 0,074(mol)`
`Na_2O+H_2O->2NaOH`
`0,074->0,074->0,148(mol)`
`a)\ C_{M\ NaOH}={0,148}/{0,2}=0,74M`
`b)\ C\%_{NaOH}={0,148.40}/{200.1,12}.100\%\approx 2,64\%`
a) Na2O + H2O → 2NaOH (1)
\(n_{Na_2O}=\dfrac{12,4}{62}=0,2\left(mol\right)\)
Theo PT1: \(n_{NaOH}=2n_{Na_2O}=2\times0,2=0,4\left(mol\right)\)
\(\Rightarrow C_{M_{NaOH}}=\dfrac{0,4}{0,4}=1\left(M\right)\)
b) 2NaOH + H2SO4 → Na2SO4 + 2H2O (2)
\(n_{NaOH}=0,2\times1=0,2\left(mol\right)\)
Theo PT2: \(n_{H_2SO_4}=\dfrac{1}{2}n_{NaOH}=\dfrac{1}{2}\times0,2=0,1\left(mol\right)\)
\(\Rightarrow V_{ddH_2SO_4}=\dfrac{0,1}{2}=0,05\left(l\right)=50ml\)
c) CO2 + 2NaOH → Na2CO3 + H2O (3)
\(n_{CO_2}=\dfrac{1,344}{22,4}=0,06\left(mol\right)\)
\(n_{NaOH}=0,1\times1=0,1\left(mol\right)\)
Theo PT: \(n_{CO_2}=\dfrac{1}{2}n_{NaOH}\)
Theo bài: \(n_{CO_2}=\dfrac{3}{5}n_{NaOH}\)
Vì \(\dfrac{3}{5}>\dfrac{1}{2}\) ⇒ CO2 dư ⇒ phản ứng tiếp
Na2CO3 + CO2 + H2O → 2NaHCO3 (4)
Theo PT3: \(n_{Na_2CO_3}=\dfrac{1}{2}n_{NaOH}=\dfrac{1}{2}\times0,1=0,05\left(mol\right)=n_{Na_2CO_3\left(4\right)}\)
\(\Rightarrow m_{Na_2CO_3}=0,05\times106=5,3\left(g\right)\)
Theo PT3: \(n_{CO_2}pư=\dfrac{1}{2}n_{NaOH}=\dfrac{1}{2}\times0,1=0,05\left(mol\right)\)
\(\Rightarrow n_{CO_2}dư=0,06-0,05=0,01\left(mol\right)=n_{CO_2\left(4\right)}\)
Theo PT4: \(n_{Na_2CO_3}=n_{CO_2}\)
Theo bài: \(n_{Na_2CO_3}=5n_{CO_2}\)
Vì \(5>1\) ⇒ Na2CO3 dư
Theo PT4: \(n_{NaHCO_3}=2n_{CO_2}=2\times0,01=0,02\left(mol\right)\)
\(\Rightarrow m_{NaHCO_3}=0,02\times84=1,68\left(g\right)\)
Theo PT4: \(n_{Na_2CO_3}pư=n_{CO_2}=0,01\left(mol\right)\)
\(\Rightarrow m_{Na_2CO_3}pư=0,01\times106=1,06\left(g\right)\)
\(\Rightarrow m_{Na_2CO_3}dư=5,3-1,06=4,24\left(g\right)\)
a) Na2O + H2O → 2NaOH (1)
nNa2O=12,462=0,2(mol)nNa2O=12,462=0,2(mol)
Theo PT1: nNaOH=2nNa2O=2×0,2=0,4(mol)nNaOH=2nNa2O=2×0,2=0,4(mol)
⇒CMNaOH=0,40,4=1(M)⇒CMNaOH=0,40,4=1(M)
b) 2NaOH + H2SO4 → Na2SO4 + 2H2O (2)
nNaOH=0,2×1=0,2(mol)nNaOH=0,2×1=0,2(mol)
Theo PT2: nH2SO4=12nNaOH=12×0,2=0,1(mol)nH2SO4=12nNaOH=12×0,2=0,1(mol)
⇒VddH2SO4=0,12=0,05(l)=50ml⇒VddH2SO4=0,12=0,05(l)=50ml
c) CO2 + 2NaOH → Na2CO3 + H2O (3)
nCO2=1,34422,4=0,06(mol)nCO2=1,34422,4=0,06(mol)
nNaOH=0,1×1=0,1(mol)nNaOH=0,1×1=0,1(mol)
Theo PT: nCO2=12nNaOHnCO2=12nNaOH
Theo bài: nCO2=35nNaOHnCO2=35nNaOH
Vì 35>1235>12 ⇒ CO2 dư ⇒ phản ứng tiếp
Na2CO3 + CO2 + H2O → 2NaHCO3 (4)
Theo PT3: nNa2CO3=12nNaOH=12×0,1=0,05(mol)=nNa2CO3(4)nNa2CO3=12nNaOH=12×0,1=0,05(mol)=nNa2CO3(4)
⇒mNa2CO3=0,05×106=5,3(g)⇒mNa2CO3=0,05×106=5,3(g)
Theo PT3: nCO2pư=12nNaOH=12×0,1=0,05(mol)nCO2pư=12nNaOH=12×0,1=0,05(mol)
⇒nCO2dư=0,06−0,05=0,01(mol)=nCO2(4)⇒nCO2dư=0,06−0,05=0,01(mol)=nCO2(4)
Theo PT4: nNa2CO3=nCO2nNa2CO3=nCO2
Theo bài: nNa2CO3=5nCO2nNa2CO3=5nCO2
Vì 5>15>1 ⇒ Na2CO3 dư
Theo PT4: nNaHCO3=2nCO2=2×0,01=0,02(mol)nNaHCO3=2nCO2=2×0,01=0,02(mol)
⇒mNaHCO3=0,02×84=1,68(g)⇒mNaHCO3=0,02×84=1,68(g)
Theo PT4: nNa2CO3pư=nCO2=0,01(mol)nNa2CO3pư=nCO2=0,01(mol)
⇒mNa2CO3pư=0,01×106=1,06(g)⇒mNa2CO3pư=0,01×106=1,06(g)
⇒mNa2CO3dư=5,3−1,06=4,24(g)