a/ PTHH: CO₂ + Ca(OH)₂ ===> CaCO₃ + H₂O

b/ n_CO2 = 2,24 / 22,4 = 0,1 mol

=> n_Ca(OH)2 = n_CO2 = 0,1 mol

=>C_M[Ca(OH)2] = 0,1 / 0,2 = 0,5M

c/ n_CaCO3 = n_CO2 = 0,1 mol

=> m_CaCO3 = 0,1 x 100 = 10 gam

nco2= 2,24/22,4=0,1 mol

co2+ ca(oh)2-> caco3+ h2o

nca(oh)2= nco2=0,1

=> cm ca(oh)2= 0,1/0,2=0,5M

ncaco3= nco2= 0,1

=>m kết tủa=0,1*100=10g

\(a.n_{CO_2}=\dfrac{0,672}{22,4}=0,03mol\\ CO_2+Ca\left(OH\right)_2\rightarrow CaCO_3+H_2O\)

\(n_{CO_2}=n_{Ca\left(OH\right)_2}=n_{CaCO_3}=0,03mol\\ m_{CaCO_3}=0,03.100=3g\\ b.V_{ddCa\left(OH\right)_2}=\dfrac{0,03}{1,5}=0,02l\)

672 mà đâu phải 0.672

\(n_{CH_4}=\dfrac{5,6}{22,4}=0,25\left(mol\right)\\ PTHH:CH_4+2O_2\underrightarrow{t^o}CO_2\uparrow+2H_2O\\ \left(mol\right)....0,25\rightarrow.......0,25\\ PTHH:CO_2+Ca\left(OH\right)_2\rightarrow CaCO_3\downarrow+H_2O\\ \left(mol\right)....0,25\rightarrow..0,25...........0,25\\ a,m_{CaCO_3}=0,25.100=25\left(g\right)\\ c,m_{Ca\left(OH\right)_2}=0,25.56=14\left(g\right)\\ C\%_{Ca\left(OH\right)_2}=\dfrac{14}{200}.100\%=7\%\)

Phương trình:

Fe₃O₄ + 4H₂SO₄ → FeSO₄ + Fe₂(SO₄)₃ + 4H₂O

FeSO₄ + 2NaOH → Na₂SO₄ + Fe(OH)₂↓

 Fe₂(SO₄)₃ + 6NaOH → 3Na₂SO₄ + 2Fe(OH)₃↓

2Fe(OH)₂ + ½ O₂ → Fe₂O₃ + 2H₂O

2Fe(OH)₃ → Fe₂O₃ + 3H₂O

Fe₂O₃ + 3CO → 2Fe + 3CO₂↑

CO₂ + Ca(OH)₂ → CaCO₃↓ + H₂O

 3CO₂ + Ca(OH)₂ → Ca(HCO₃)₂

Ca(HCO₃)₂ → CaCO₃ + CO₂↑ + H₂O

a/ Ca(OH)2 + CO2--> CaCO3 + H2O

nCO2=3.36/22.4=0.15(mol)

nCO2=nCa(OH)2=0.15(mol)

200ml=0.2l

b/CMCa(OH)2= 0.15/0.2=0.75(M)

c/nCaCO3=nCO2=0.15(mol)

mCaCO3= 0.15 x 106=15.9(g)

\(a.n_{CO_2}=\dfrac{5,6}{22,4}=0,25\left(mol\right)\\ a.Ca\left(OH\right)_2+CO_2\rightarrow CaCO_3+H_2O\\ 0,25.......0,25............0,25..........0,25\left(mol\right)\\ C_{MddCa\left(OH\right)_2}=\dfrac{0,25}{0,1}=2,5\left(M\right)\\ b.m_{\downarrow}=m_{CaCO_3}=100.0,25=25\left(g\right)\)