\(Ca\left(OH\right)_2+CO_2\rightarrow CaCO_3\downarrow+H_2O\left(1\right)\\ Ca\left(OH\right)_2+2CO_2\rightarrow Ca\left(HCO_3\right)_2\left(2\right)\\ Ca\left(HCO_3\right)_2\rightarrow CaCO_3\downarrow+CO_2\uparrow+H_2O\left(3\right)\)

\(n_{CaCO_3\left(1\right)}=\dfrac{15}{100}=0,15\left(mol\right)\)

\(n_{Ca\left(HCO_3\right)_2}=n_{CaCO_3\left(3\right)}=\dfrac{2,5}{100}=0,025\left(mol\right)\)

\(\left(1\right)\left(2\right)\left(3\right)\Rightarrow\sum n_{CO_2}=0,15+2.0,025=0,2\left(mol\right)\\ \Rightarrow V=22,4.0,2=4,48\left(l\right)\)

CO2 + Ca(OH)2 thu được kết tủa và dung dịch X

=> \(n_{CaCO_3}=\dfrac{15}{100}=0,15\left(mol\right)\)

Đun X lại thu được kết tủa

=> X có Ca(HCO3)2

=> \(n_{kt}=n_{Ca\left(HCO_3\right)_2}=\dfrac{2,5}{100}=0,025\left(mol\right)\)

Bảo toàn nguyên tố C : \(n_{CO_2}=0,15+0,025.2=0,2\left(mol\right)\)

=> \(V_{CO_2}=0,2.22,4=4,48\left(l\right)\)

Đáp án D

n_C4H10 = 0,56/22,4 = 0,025 mol

=> n_CO2 = 4.0,025 = 0,1 mol; n_H2O = 5.0,025 = 0,125 mol;

T = n_OH-/n_CO2 = 0,16/0,1 = 1,6

 => tạo ra 2 muối; n_CO3^2- = 0,06 => m_CaCO3 = 0,06.197 = 11,82 gam

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)