a)Quy \(\left\{{}\begin{matrix}Na:x\left(mol\right)\\Ba:y\left(môl\right)\\O:z\left(mol\right)\end{matrix}\right.\underrightarrow{+H_2O}\left\{{}\begin{matrix}NaOH:x\left(mol\right)\\Ba\left(OH\right)_2:y\left(mol\right)\\O^{2-}:z\left(mol\right)\end{matrix}\right.+H_2\)

\(n_{H_2}=\dfrac{1,12}{22,4}=0,05mol\)

\(n_{Ba\left(OH\right)_2}=\dfrac{20,52}{171}=0,12mol\Rightarrow y=0,12mol\)

Ta có hệ: \(\left\{{}\begin{matrix}BTKL:23x+137y+16z=21,9\\y=0,12\\BTe:x+2y=2z+2n_{H_2}\Rightarrow x-2z=-0,14\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x=0,14\\y=0,12\\z=0,14\end{matrix}\right.\)

\(n_{OH^-}=n_{NaOH}+2n_{Ba\left(OH\right)_2}=0,14+2\cdot0,12=0,38mol\)

\(n_{CO _2}=\dfrac{6,72}{22,4}=0,3mol\Rightarrow n_{CO_3^{2-}}=0,38-0,3=0,08mol\)

\(\Rightarrow m_{CO_3^{2-}\downarrow}=0,08\cdot197=15,76g\)

Phần c thì làm ntn ạ

2/
Hỗn hợp X gồm 1 andehit, 1 axit, 1 este khi đốt cho nCO2 = nH2O => các chất trong X đều no, đơn chức.
CnH2nOx ----> nCO2
a.........................4a
=> n = 4
=> Este và acid có CT: C4H8O2, Andehit có CT: C4H8O
Khi X tác dụng với với dd NaOH thì chỉ có acid và este pứ => n este = n ancol = 0.5a
=> n acid = 0.75a - 0.5a = 0.25a
=> n andehit = 0.25a
=> %m acid = 0.25*88/(0.25*88 + 0.25*72 + 0.5*88) = 26.19%

a, \(n_{H_2}=\dfrac{1,12}{22,4}=0,05\left(mol\right)\)

PT: \(2Na+2H_2O\rightarrow2NaOH+H_2\)

\(Na_2O+H_2O\rightarrow2NaOH\)

Theo PT: \(n_{Na}=2n_{H_2}=0,1\left(mol\right)\Rightarrow m_{Na}=0,1.23=2,3\left(g\right)\)

\(\Rightarrow m_{Na_2O}=8,5-2,3=6,2\left(g\right)\)

b, \(n_{Na_2O}=\dfrac{6,2}{62}=0,1\left(mol\right)\)

Theo PT: \(n_{NaOH}=n_{Na}+2n_{Na_2O}=0,3\left(mol\right)\Rightarrow m_{NaOH}=0,3.40=12\left(g\right)\)

PTHH: \(Na+H_2O\rightarrow NaOH+\dfrac{1}{2}H_2\uparrow\)  (1)

           \(Na_2O+H_2O\rightarrow2NaOH\)  (2)

           \(Fe+H_2SO_4\rightarrow FeSO_4+H_2\uparrow\)  (3)

Ta có: \(\left\{{}\begin{matrix}n_{Na}=2n_{H_2\left(1\right)}=2\cdot\dfrac{2,24}{22,4}=0,2\left(mol\right)\\n_{Fe}=n_{H_2\left(3\right)}=\dfrac{1,12}{22,4}=0,05\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}\%m_{Na}=\dfrac{0,2\cdot23}{16,4}\cdot100\%\approx28,05\%\\\%m_{Fe}=\dfrac{0,05\cdot56}{16,4}\cdot100\%\approx17,07\%\\\%m_{Na_2O}=54,88\%\end{matrix}\right.\)

bạn ơi 16,4 bạn lấy đâu ra vậy???

a)

\(n_{H_2}=\dfrac{0,336}{22,4}=0,015\left(mol\right)\)

PTHH: 2Na + 2H₂O --> 2NaOH + H₂

          0,03<------------0,03<----0,015

=> \(\%m_{Na}=\dfrac{0,03.23}{1,31}.100\%=52,67\%\)

=> \(\%m_{Na_2O}=100\%-52,67\%=47,33\%\)

b)

\(n_{Na_2O}=\dfrac{1,31.47,33\%}{62}=0,01\left(mol\right)\)

PTHH: Na₂O + H₂O --> 2NaOH

            0,01----------->0,02

=> n_NaOH = 0,03 + 0,02 = 0,05 (mol)

m_{dd sau pư} = 1,31 + 18,72 - 0,015.2 = 20 (g)

=> \(C\%_{dd.NaOH}=\dfrac{0,05.40}{20}.100\%=10\%\)

\(V_{dd.NaOH}=\dfrac{20}{1,2}=\dfrac{50}{3}\left(ml\right)=\dfrac{1}{60}\left(l\right)\) 

\(C_{M\left(dd.NaOH\right)}=\dfrac{0,05}{\dfrac{1}{60}}=3M\)

\(n_{Na_2O}=\dfrac{a}{62}\left(mol\right)\)

\(m_{dd}=a+b\left(g\right)\)

`Na_2O + H_2O -> 2NaOH`

Theo PT: `n_{NaOH} = 2n_{Na_2O} = a/(31) (mol)`

\(\Rightarrow C\%_{NaOH}=\dfrac{40.\dfrac{a}{31}}{a+b}.100\%=10\%\)

\(\Leftrightarrow\dfrac{40a}{31}=0,1\left(a+b\right)\\ \Leftrightarrow\dfrac{369a}{310}=0,1b\\ \Leftrightarrow\dfrac{a}{b}=\dfrac{0,1}{\dfrac{369}{310}}=\dfrac{31}{369}\)

Hay `a : b = 31 : 369`

\(Na_2O+H_2O->2NaOH\\ C\%=0,1=\dfrac{\dfrac{a}{62}\cdot2\cdot40}{a+b}\\ 0,1a+0,1b=\dfrac{40}{31}a\\ 0,1b=\dfrac{369}{310}a\\ a:b=\dfrac{0,1}{\dfrac{369}{310}}=\dfrac{31}{369}=0,08401\)