4Na+O2----->2Na2O

Na2O+H2O--->2NaOH

2NaOH dư+CO2----->Na2CO3+H2O

Na2CO3+CaCl2------>CaCO3+2NaCl

CaCO3---t0---->CaO+CO2

CaO+H2O------>Ca(OH)2

Ca(OH)2+CO2 dư----->Ca(HCO3)2

-)2Na+2H2O----->2NaOH+H2

 FeO+2NaOH----->Na2O+Fe(OH)2

Na2O+CO2------->Na2CO3

$\Rightarrow \begin{array}{c}2x+y=1,2\end{array}$

  x+y=0,75
$\Rightarrow \begin{array}{c}x=0,\mathrm{45;}\\ y=0,3\end{array}$

$m_{N{a}_{2}C{O}_3}$=n.M=0,3.84=25,2(g)
$m_{NaHC{O}_3}$=n.M=0,45.106=47,7(g)
$m_A$=25,2+47,7=72,9(g)

b,$N{a}_{2}C{O}_3+BaC{l}_2-->BaC{O}_3+2NaCl$
0.45--------------------->0.45mol
m$BaC{O}_3$=197.0.45=88.65g

tạo 1 pt hoàn chỉnh ấy à!!!

\(n_{CO2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\)

Pt : \(CO_2+2NaOH\rightarrow Na_2CO_3+H_2O|\)

         1            2                   1              1

       0,3         0,6

\(n_{NaOH}=\dfrac{0,3.2}{1}=0,6\left(mol\right)\)

200ml = 0,2l

\(C_{M_{ddNaOH}}=\dfrac{0,6}{0,2}=3\left(M\right)\)

 Chúc bạn học tốt

a) $C_2H_5OH + 3O_2 \xrightarrow{t^o} 2CO_2 + 3H_2O$

b) $n_{C_2H_5OH} = \dfrac{9,2}{46} = 0,2(mol)$

Theo PTHH : 

$n_{CO_2} = 2n_{C_2H_5OH} = 0,4(mol) \Rightarrow V_{CO_2} = 0,4.22,4 = 8,96(lít)$

$n_{H_2O} = 3n_{C_2H_5OH} = 0,6(mol) \Rightarrow m_{H_2O} = 0,6.18 = 10,8(gam)$

c) $CO_2 + Ca(OH)_2 \to CaCO_3 + H_2O$
$n_{CaCO_3} = n_{CO_2} = 0,4(mol)$

$m_{CaCO_3} = 0,4.100 = 40(gam)$

\(n_{NaOH}=0,3.1=0,3\left(mol\right)\\ n_{CO_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\\1< \dfrac{n_{NaOH}}{n_{CO_2}}=\dfrac{0,3}{0,2}=1,5< 2\)

=> Sản phẩm tạo 2 muối: Na₂CO₃ và NaHCO₃

=> CHỌN A

A

chọn A

`n_[Na_2 CO_3]=[2,76]/106=0,03(mol)`

`Na_2 CO_3 +2CH_3 COOH->2CH_3 COONa+H_2 O+CO_2\uparrow`

   `0,03`                    `0,06`                                                           `0,03`             `(mol)`

`CO_2 +Ca(OH)_2 ->CaCO_3 \downarrow+H_2 O`

 `0,03`                                  `0,03`

`a)CH_3 COONa` là muối natri axetat.

  `V_[dd CH_3 COOH]=[0,06]/[0,2]=0,3(l)`

`b)m_[CaCO_3]=0,03.100=3(g)`

a, CH₃COONa: Natri axetat

PT: \(Na_2CO_3+2CH_3COOH\rightarrow2CH_3COONa+CO_2+H_2O\)

Ta có: \(n_{Na_2CO_3}=\dfrac{2,76}{106}=0,026\left(mol\right)\)

Theo PT: \(n_{CH_3COOH}=2n_{Na_2CO_3}=0,052\left(mol\right)\Rightarrow V_{CH_3COOH}=\dfrac{0,052}{0,2}=0,26\left(l\right)\)

b, \(CO_2+Ca\left(OH\right)_2\rightarrow CaCO_{3\downarrow}+H_2O\)

Theo PT: \(n_{CaCO_3}=n_{CO_2}=n_{Na_2CO_3}=0,026\left(mol\right)\)

\(\Rightarrow m_{CaCO_3}=0,026.100=2,6\left(g\right)\)

a.b.\(n_{CO_2}=\dfrac{5,6}{22,4}=0,25mol\)

\(2CH_3COOH+Na_2CO_3\rightarrow2CH_3COONa+CO_2+H_2O\)

       0,5                                                              0,25               ( mol )

\(m_{CH_3COOH}=0,5.60=30g\)

\(\%m_{CH_3COOH}=\dfrac{30}{45}.100=66,67\%\)

\(\%m_{C_2H_5OH}=100\%-66,67\%=33,33\%\)

c.\(m_{NaOH}=50.20\%=10g\)

\(n_{NaOH}=\dfrac{10}{40}=0,25mol\)

\(2NaOH+CO_2\rightarrow Na_2CO_3+H_2O\)

 \(\dfrac{0,25}{2}\)   <    \(\dfrac{0,25}{1}\)                                ( mol )

 0,25                           0,125                    ( mol )

\(m_{Na_2CO_3}=0,125.106=13,25g\)

a) 

2CH₃COOH + Na₂CO₃ --> 2CH₃COONa + CO₂ + H₂O

b) \(n_{CO_2}=\dfrac{5,6}{22,4}=0,25\left(mol\right)\)

PTHH: 2CH₃COOH + Na₂CO₃ --> 2CH₃COONa + CO₂ + H₂O

                  0,5<-----------------------------------0,25

=> m_CH3COOH = 0,5.60 = 30 (g)

=> m_C2H5OH = 45 - 30 = 15 (g)

c) \(n_{NaOH}=\dfrac{50.20\%}{40}=0,25\left(mol\right)\)

Xét tỉ lệ: \(\dfrac{n_{NaOH}}{n_{CO_2}}=\dfrac{0,25}{0,25}=1\) => Tạo muối NaHCO₃

PTHH: NaOH + CO₂ --> NaHCO₃

              0,25-------------->0,25

=> m_NaHCO3 = 0,25.84 = 21 (g)

Đáp án B