CaO+ H₂O ------> Ca(OH)₂

0.1......0.1..................0.1

n_CaO=0.1 mol

a) n_CO2=0.15 mol

Xét tỉ lệ: 1<\(\dfrac{n_{CO2}}{n_{Ca\left(OH\right)2}}=\dfrac{0.15}{0.1}=1.5\)< 2

Do đó hh muối gồm CaCO₃ và aCHCO₃

Ca(OH)₂+ CO₂ -------> CaCO₃+ H₂O

a................a......................a...........a

Ca(OH)₂+ 2CO₂ -------> Ca(HCO₃)₂

b................2b........................b

Gọi a, b là số mol CaCO₃ và Ca(HCO₃)₂

Ta có hệ: \(\left\{{}\begin{matrix}n_{Ca\left(OH\right)2}=a+b=0.1\\n_{CO2}=a+2b=0.15\end{matrix}\right.\Rightarrow a=b=0.05\)

=>m_CaCO3=100a=5g

b) Có 2 trường hợp:

+ TH₁:CO₂ lấy thiếu

Ca(OH)₂+ CO₂ --------> CaCO₃↓+ H₂O

0.02..........0.02...................0.02......0.02

n_CaCO3=0.02 mol

=> V_CO2=0.02* 22.4=0.448 lít

+ TH₂: CO₂ lấy dư

Ca(OH)₂+ CO₂ ------> CaCO₃+ H₂O

0.02...........0.02..............0.02.......0.02

Ca(OH)₂+ 2CO₂ -----> Ca(HCO₃)₂

0.08...........0.16..................0.08

n_CaCO3=0.02 mol

n_Ca(OH)2=0.1-0.02=0.08 mol

=> V_CO2=(0.02+0.16)*22.4=4.032 lít

CHÚC BẠN HỌC TỐT!!

a) Theo đề bài, ta có: \(\left\{{}\begin{matrix}n_{CO2}=\dfrac{0,448}{22,4}=0,02\left(mol\right)\\n_{Ca\left(OH\right)2}=0,02.0,4=0,008\left(mol\right)\end{matrix}\right.\)

Ta có tỉ lệ: \(\dfrac{n_{Ca\left(OH\right)2}}{n_{CO2}}=\dfrac{0,008}{0,02}=0,4\)

Vì 0,4 < 1 nên sản phẩm tạo thành là \(Ca\left(HCO_3\right)_2\)

_{PTHH: \(Ca\left(OH\right)_2+2CO_2\rightarrow Ca\left(HCO_3\right)_2\)}

pư.........0,008........0,016.......0,008 (mol)

\(\Rightarrow\left\{{}\begin{matrix}C_{M\left(CaCO3\right)}=\dfrac{0,008}{0,448+0,4}\approx9,43\left(M\right)\\C_{MCa\left(OH\right)2dư}=\dfrac{0,02-0,008}{0,448+0,4}\approx0,01\left(M\right)\end{matrix}\right.\)

Vậy..............

CHÚC BẠN HỌC TỐT!!

b) \(\left\{{}\begin{matrix}n_{SO2}=\dfrac{4,032}{22,4}=0,18\left(mol\right)\\n_{Ba\left(OH\right)2}=1.0,2=0,2\left(mol\right)\end{matrix}\right.\)

Ta có tỉ lệ: \(\dfrac{n_{Ba\left(OH\right)2}}{n_{SO2}}=\dfrac{0,2}{0,18}=\dfrac{10}{9}\)

Vì \(\dfrac{10}{9}>1\) nên sản phẩm tạo thành là: \(Ba\left(HSO_3\right)_2\)

PTHH: \(Ba\left(OH\right)_2+SO_2\rightarrow BaSO_3\downarrow+H_2O\)

pư................0,18.........0,18............0,18..........0,18 (mol)

\(\Rightarrow\left\{{}\begin{matrix}C_{MBaCO3}=\dfrac{0,18}{0,2}=0,9\left(M\right)\\C_{MBa\left(OH\right)2dư}=\dfrac{0,2-0,18}{0,2}=0,1\left(M\right)\end{matrix}\right.\)

\(n_{CO_2}=\dfrac{2.24}{22.4}=0.1\left(mol\right)\)

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

\(0.1.............0.1\)

\(C_{M_{Ba\left(OH\right)_2}}=\dfrac{0.1}{0.1}=1\left(M\right)\)

\(n_{CO_2}=\dfrac{2,24}{22,4}=0,1(mol)\\ PTHH:CO_2+Ba(OH)_2\to BaCO_3\downarrow+H_2O\\ \Rightarrow n_{Ba(OH)_2}=n_{CO_2}=0,1(mol)\\ \Rightarrow C_{M_{Ba(OH)_2}}=\dfrac{0,1}{0,1}=1M\)

\(n_{CO_2}=0,05mol\)

\(\dfrac{n_{CO_2}}{n_{Ca\left(OH\right)_2}}=\dfrac{0,05}{0,2}=0,25< 1\)

Ca(OH)₂ dư tạo muối CaCO₃

CO₂+Ca(OH)₂\(\rightarrow\)CaCO₃+H₂O

\(n_{Ca\left(OH\right)_2}=n_{CaCO_3}=n_{CO_2}=0,05mol\)

\(m_{CaCO_3}=0,05.100=5g\)

\(n_{Ca\left(OH\right)_2}\left(dư\right)=0,2-0,05=0,15mol\)

\(m_{Ca\left(OH\right)_2}=0,15.74=3,7g\)

a.

nH2S=17.92/22.4=0.8(mol)

nNaOH=1.26*1=1.26(mol)

H₂S+2NaOH→2H₂O+Na₂S

nH2O=nNaOH=1.26(mol)=>mH2O=1.26*18=22.68(gam)

nNa2S=nH2S=0.8(mol)=>mNa2S=0.8*78=62.4(gam)

b.

nH2S=6.72/22.4=0.3(mol)

nKOH=0.15*2=0.3(mol)

H₂S+2KOH→2H₂O+K₂S

nH2O=nKOH=0.3(mol)=>mH2O=0.3*18=5.4(gam)

nK2S=nH2S=0.3(mol)=>mK2S=0.3*78=23.4(gam)

c.

nH2S=2.24/22.4=0.1(mol)

nNaOH=0.3*1=0.3(mol)

H₂S+2NaOH→2H₂O+Na₂S

nH2O=nNaOH=0.1(mol)=>mH2O=0.1*18=1.8(gam)

nNa2S=nH2S=0.3(mol)=>mNa2S=0.3*78=23.4(gam)

d.

nSO2=12.8/64=0.2(mol)

nNaOH=0.25*1=0.25(mol)

2NaOH+SO₂→H₂O+Na₂SO₃

nH2O=nSO2=0.2(mol)=>mH2O=0.2*18=3.6(gam)

nNa2SO3=nSO2=0.2(mol)=>mNa2SO3=0.2*126=25.2(gam)

e.

nSO2=13.44/64=0.21(mol)

nNaOH=0.2*2=0.4(mol)

2NaOH+SO₂→H₂O+Na₂SO₃

nH2O=nSO2=0.21(mol)=>mH2O=0.21*18=3.78(gam)

nNa2SO3=nSO2=0.4(mol)=>mNa2SO3=0.4*126=50.4(gam)

f.

nH2S=5.6/22.4=0.25(mol)

nKOH=0.4*1.5=0.6(mol)

H₂S+2KOH→2H₂O+K₂S

nH2O=nKOH=0.25(mol)=>mH2O=0.25*18=4.5(gam)

nK2S=nH2S=0.6(mol)=>mK2S=0.6*78=46.8(gam)