nCO2=0,15mol

nNa2CO3=0,25mol

PTHH: CO2+2NaOH=>Na2CO3 + H2O

            0,15:                       0,25    =>nNa2CO3 dư

p/ư:        0,15----0,3---------->0,15

=>V=0,3:1,2=0.25lit

=>  CM(Na2CO3)=0,15:0,25=0,6M

Câu 1:

\(CO_2+Ba\left(OH\right)_2\rightarrow BaCO_3+H_2O\)

x_________x _______x ___________

\(2CO_2+Ba\left(OH\right)_2\rightarrow Ba\left(HCO_3\right)_2\)

y_________y/2____________

\(x+\frac{y}{2}=0,75\left(1\right)\)

\(klg_{dd\left(giam\right)}=klg_{kt}-klg_{CO2}\)

\(\Rightarrow5,45=197x-\left(x+y\right).44\left(2\right)\)

\(\left(1\right)+\left(2\right)\Rightarrow\left\{{}\begin{matrix}x=0,3\\y=0,9\end{matrix}\right.\)

\(\Rightarrow V=26,88\left(l\right)\)

Câu 2:

\(n_{SO2}=n_S=0,05\left(mol\right)\)

Cho 0,05mol SO2 vào 0,2 mol OH^- thì tạo thành 0,05mol CO₃^2-

^{\(\Rightarrow m_{\downarrow}=217.0,05=10,85\left(g\right)\)}

a, VddHCl=0,42,5=0,16(l)VddHCl=0,42,5=0,16(l)

45.

\(n_{SO2}=\frac{38,528}{22,4}=1,72\left(mol\right)\)

\(m_{NaOH}=\frac{2832.100}{100}=283,2\left(g\right)\)

\(\Rightarrow n_{NaOH}=283,2:40=7,08\left(mol\right)\)

Tỉ lệ \(\frac{n_{NaOH}}{n_{SO2}}=4,12>2\Rightarrow\) Na2SO3 hết , NaOH dư

\(PTHH:SO_2+NaOH\rightarrow Na_2SO_3+H_2O\)

______1,72______1,72___________________

\(\Rightarrow m_{Na2SO3}=\frac{1,72}{126}=216,72\left(g\right)\)

46.

\(CO_2+Ca\left(OH\right)_2\rightarrow CaCO_3+H_2O\)

0,1_______________0,1 _________(mol)

\(n_{Ca\left(OH\right)2}=0,5.0,3=0,15\left(mol\right)\)

\(n_{CaCO3}=\frac{10}{100}=0,1\left(mol\right)\)

Tỉ lệ : \(\frac{0,15}{1}>\frac{0,1}{1}\Rightarrow\) nCa(OH)2 phản ứng dư, nCaCO3 phản ứng hết

\(\Rightarrow V_{CO2}=0,1.22,4=2,24\left(l\right)\)

47.

\(CaO+H_2O\rightarrow Ca\left(OH\right)_2\)

\(\Rightarrow n_{Ca\left(OH\right)2}=n_{CaO}=0,1\left(mol\right)\)

\(n_{CaSO3}=0,01\left(mol\right)\)

- TH1: Dư kiềm

\(Ca\left(OH\right)_2+SO_2\rightarrow CaSO_3+H_2O\)

\(\Rightarrow n_{SO2}=n_{CaSO3}=0,01\left(mol\right)\)

\(\Rightarrow V=0,01.22,4=0,224\left(l\right)\)

- TH2: Dư SO2

\(Ca\left(OH\right)_2+SO_2\rightarrow CaSO_3+H_2O\)

\(\Rightarrow n_{SO2_{kt}}=n_{Ca\left(OH\right)2}=n_{CaSO3}=0,1\left(mol\right)\)

Vậy Có 0,1- 0,01 = 0,09 mol CaSO3 tan

\(CaSO_3+SO_2+H_2O\rightarrow Ca\left(HSO_3\right)_2\)

\(\Rightarrow n_{SO2_{kt}}=0,09\left(mol\right)\)

Tổng mol SO2= 0,19 mol

\(\Rightarrow V=0,19.22,4=4,256\left(l\right)\)

48.

\(n_{NaOH}=0,781\left(mol\right)\)

Gọi a là mol NaHSO3, b là mol Na2SO3

\(\Rightarrow104a+126b=79,338\left(1\right)\)

\(NaOH+SO_2\rightarrow NaHSO_3\)

\(2NaOH+SO_2\rightarrow Na_2SO_3+H_2O\)

\(\Rightarrow a+2b=0,781\left(2\right)\)

\(\left(1\right)+\left(2\right)\Rightarrow\left\{{}\begin{matrix}a=0,735\\b=0,023\end{matrix}\right.\)

\(m_{NaHSO3}=104.0,735=76,44\left(g\right)\)

\(m_{Na2SO3}=126.0,023=2,898\left(g\right)\)

ai trả lời giúp mình vs mình đg cần gấp

\(n_{H_2S}=\dfrac{0.672}{22.4}=0.03\left(mol\right)\)

\(a.\)

\(n_{NaOH}=0.75\cdot0.1=0.075\left(mol\right)\)

\(T=\dfrac{0.075}{0.03}=2.5>2\)

=> Tạo muối trung hòa

\(2NaOH+H_2S\rightarrow Na_2S+H_2O\)

\(0.06............0.03.........0.03\)

\(m_{Na_2S}=0.03\cdot78=2.34\left(g\right)\)

\(b.\)

\(n_{NaOH}=0.42\cdot0.1=0.042\left(mol\right)\)

\(T=\dfrac{0.042}{0.03}=1.4\)

=> Tạo 2 muối

\(n_{Na_2S}=a\left(mol\right),n_{NaHS}=b\left(mol\right)\)

\(\left\{{}\begin{matrix}2a+b=0.042\\a+b=0.03\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}a=0.012\\b=0.018\end{matrix}\right.\)

\(m_{Muối}=0.012\cdot78+0.018\cdot56=1.944\left(g\right)\)

a, \(n_{CO_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\)

PTHH: Na₂CO₃ + 2HCl → NaCl + CO₂ + H₂O

Mol:        0,2            0,4         0,2      0,2

\(m_{Na_2CO_3}=0,2.106=21,2\left(g\right)\)

\(V_{ddHCl}=\dfrac{0,4}{2,5}=0,16\left(l\right)\)

\(C_{M_{ddNaCl}}=\dfrac{0,2}{0,16}=1,25M\)

`C_2H_4+3O_2 to->2CO_2+2H_2O`

0,04------------------------0,08 mol

`Ca(OH)_2+ CO_2->CaCO_3+H_2O`

0,08--------------0,08----------0,08 mol

`m_(CaCO_3)=8/100=0,08 mol`

`V_(C_2H_4)=0,04.22,4=0,896l`

b) Sau thí nghiệm bình tăng :

`m_ bình tăng=0,08.18=1,44g`